Symmetry and the Monster (7): First the Medical Report, Then the Monster

[Author’s note — Yeqiu]

When we ask whether an AI’s proof is reliable, behind that question there is in fact a bigger one: why is a proof that we have not verified ourselves reliable? Our usual answer is probably something like peer review, verifiability and so on. But when we measure AI’s proofs by the same standard, we run into a question that cannot be avoided: is AI a peer? Or rather, is AI qualified to be a peer?

And the word “qualified” may be heavier than it looks. A peer is not just someone capable enough; a peer is someone who makes mistakes and has to answer for their own mistakes: staking a reputation, staking a career. AI, by contrast, has nothing it can lose. This chain of questions is probably hard to answer to this day even for the top scientists in the field. But it is not far away; it is a question our present era is facing, and it is closely bound up with how we actually use AI.

I don’t have an answer. But it is exactly this question without an answer that pushes me toward another, inward-looking question: in such an era, how should a person conduct themselves?

In the author’s notes of earlier installments I mentioned that in the AI era, reading books will become more important, not less. Because in this era, reading is a personal experience, and it is also the foundation on which a person forms their own thought. And personal experience will become more and more important, because AI is changing this era very fast; if we still measure people and AI by the old standards, such as how much code one has written, how much knowledge one has stored, how efficiently one completes tasks, then people are bound to be at a disadvantage.

But what is truly unique and irreplaceable about people comes precisely from our personal experience, from the experience, thought and feeling formed by that experience; if they benefit another person or let another person feel the same, that moment is happiness.

Episode 7 of the series. In the previous episode we opened the box holding the 26 sporadic simple groups, watched them come on stage one after another across several decades, and said: all these sporadic groups in the end point faintly toward the same monster, unreasonably large. In this episode we look at this monster head-on (it has a formal name, the Monster): how it was first “computed” and then “built” by one person by hand. It lives in 196883 dimensions. And why does 196884 (exactly 1 more than 196883) appear almost unchanged in a seemingly unrelated corner of number theory? That crack is left for the last episode.

“317 is a prime, not because we think so, or because our minds are shaped in one way rather than another, but because it is so, because mathematical reality is built that way.”

— G. H. Hardy (1877–1947), A Mathematician’s Apology §24

Companion video (YouTube)

Prelude · First the medical report, then the thing itself

There is a rare thing in the history of mathematics, and it happened precisely to the protagonist of this episode: before anyone saw this monster, mathematicians had already written its medical report: how big it is, what its “fingerprint table” looks like, the minimum number of dimensions of the space it has to live in, all of it had been computed. Then several more years passed before someone actually built it, confirmed its identity, and matched it point by point against the report written in advance.

An object’s medical report existing before the object itself: in other fields this is almost unimaginable. It is hard to measure an animal’s height and weight first and then go back and discover the animal. But this is exactly how the Monster came on stage: first the numbers, then the thing. What this episode tells is how this report was computed (§1, §2), how the person who really put the monster up completed it by hand (§3), and finally we take stock: is there anything else left in the box at all (§4)?

§1 · Fischer’s mirrors

One end of the story is in Germany in the 1960s. Bernd Fischer (1936–2020), a group theorist who later led mathematics at the University of Bielefeld, asked a question that looked too plain to lead to a monster.

Bernd Fischer (1936–2020): his plain question, “mirrors may only meet at $90^\circ$ or $60^\circ$”, led in the end to the Monster. (Photo by Katrin Breithaupt, 2008, Oberwolfach Photo Collection, CC BY-SA 2.0 de.)

To explain this question clearly, we first have to talk about the person who led him onto this road. In secondary school Fischer had an excellent mathematics teacher, who before World War II had spent three years as an assistant in Darmstadt computing rocket trajectories (that was exactly the era in which Germany led in rockets; the V2 came out of the late phase of World War II), with real, hands-on differential equations. But this teacher was not a rocket expert; he was a mathematician; and what really moved Fischer was not the rockets, but his way of using mathematics to handle physical problems. How good was he? After Fischer went to university, he did not need to attend a single lecture of the differential equations course. Influenced by this teacher, his plan on entering university was to do a master’s in physics first and then a doctorate in mathematics; in other words, he had originally meant to go into applied mathematics.

Reinhold Baer (1902–1979): the teacher who led Fischer from “would-be applied mathematician” to pure mathematician; when teaching analysis he would deliberately make mistakes and wait for the students to correct him. (Photo by Konrad Jacobs, Oberwolfach Photo Collection, CC BY-SA 2.0 de.)

But the one who led him toward another road was Reinhold Baer, a professor who had just returned to Germany from the United States. What fascinated Fischer was not the content of some particular course, but Baer’s way of doing mathematics, and his way of talking to these young students. The example Fischer gives is: when Baer taught analysis he would deliberately make mistakes and wait for the students to correct him, an act that made “you are entitled to question me” an everyday thing. He admired Baer for two more things: first, breadth, wanting to bring every direction into the seminar; second, freedom: you could do whatever you wanted, which was very unusual in German universities at the time. These things together turned a would-be applied mathematician into a pure mathematician. Baer’s seminar also drew almost everyone in this field to Frankfurt: Tits (Jacques Tits, 1930–2021) came often, Thompson (John Thompson, 1932–) came, and Janko (Zvonimir Janko, 1932–2022) came too.

And Fischer himself recalled that one of the things he read after burrowing into the library as a student was a structure called a “distributive quasigroup”; he felt at the time that there was plainly a group hidden behind it. This later became exactly the topic of his doctoral thesis, and exactly the starting point that led him onto the road of “groups generated by mirror-like operations”.

Fischer’s question. Imagine a batch of operations, each one like a mirror: reflect once, left and right are flipped; reflect twice, back to how it was (such an element that “returns to the original after multiplying by itself once” is called an involution in mathematics; we already dealt with it in EP5). What Fischer wanted to study were groups generated by a whole class of such “mirrors”; but he added a very tight restriction: take any two mirrors and reflect in them one after the other; the order of this composite operation may only be 1, 2 or 3, nothing larger.

Translated into geometry, this restriction is a very clean sentence. Reflecting in two mirrors one after the other composes to a rotation, whose angle is exactly twice the angle between the two mirrors. So “the order of the composite operation is at most 3” is the same as saying: any two mirrors either coincide, or meet at 90°, or meet at 60°, and no other angle is allowed. The whole group then becomes a huge network of mirrors, and what Fischer asked was: how many kinds of such networks can there be in all?

The answer is abnormally short. Fischer proved around 1970: the “atoms of symmetry” (which can be understood as the simple groups that cannot be divided further) satisfying these rules come in exactly six classes. Five of them are long-familiar faces: all rearrangements $S_n$; three families of classical groups over $\mathbb{F}_2$ (the smallest field, with only the two elements $0$ and $1$): the symplectic groups $\mathrm{Sp}_{2n}(2)$, the unitary groups $\mathrm{PSU}_n(2)$, the orthogonal groups $\mathrm{O}^{\pm}_{2n}(2)$ (these three families all grow on the Killing–Cartan periodic table of EP4, and the Chevalley recipe told in EP5 §1 is the procedure that carries them over from the real numbers to finite fields: the symplectic groups are the $C$ family on the table, which is what EP4 called them; the orthogonal groups are the $B$/$D$ families, which EP4 wrote as $SO(k)$ on the real-number side; the unitary groups are a branch twisted out of the $A$ family, which EP5 Appendix D even constructed explicitly); plus one family of orthogonal-type groups over $\mathbb{F}_3$. And the sixth class is three brand-new sporadic groups that nobody had seen before, today written $\mathrm{Fi}_{22}$, $\mathrm{Fi}_{23}$, $\mathrm{Fi}_{24}'$. (For the structure of this classification, and an example you can compute by hand ($S_n$ itself is a network of mirrors), see Appendix A.)

A warning bell has to be struck here, one already planted at the end of EP6: Fischer’s $\mathrm{Fi}$ and Mathieu’s $M$ from the previous episode are two different families. Fischer, to save trouble back then, wrote these three groups as $M(22)$, $M(23)$, $M(24)$, deliberately echoing Mathieu; only when the big atlas was published in 1985 were they uniformly changed to $\mathrm{Fi}$ (the first two letters of Fischer), to avoid a name clash with the Mathieu groups. The same subscripts, completely different groups, and sizes that differ by an enormous distance: the order of Mathieu’s $M_{22}$ is 443,520, while Fischer’s $\mathrm{Fi}_{22}$ is about $6.5\times10^{13}$, about one hundred fifty million times larger.

Loosen the mirror rule by one notch, and you are on the road to the Monster. After Fischer stepped down as dean and returned to research, he loosened the restriction from “angles may only be 90°/60°” to also allowing 45° (corresponding to composite operations of order up to 4). Why 4, and not larger? There is no limit given by mathematics here: that was the step Fischer took at the time, and it succeeded. Pushing the same idea of mirror networks further, in the summer of 1973 he caught the largest group among them, which Conway (John Conway, 1937–2020) later named the Baby Monster. How big is it? Its order is

$|B| = 4{,}154{,}781{,}481{,}226{,}426{,}191{,}177{,}580{,}544{,}000{,}000 \approx 4.15\times10^{33}.$

About this number there is a true little episode: Fischer was visiting Cambridge, and Conway wanted to use an old mechanical calculator to help him compute the last step, but could not find the missing part however he looked, and in the end Fischer and his wife simply finished it by hand. Fischer said he had something to go on: he knew the answer had to be divisible by 31, so the computation came with a built-in check. In August of that year, at the meeting in Oberwolfach, he spoke publicly about this new group for the first time.

The lecture register of this meeting still survives: the Oberwolfach institute’s Vortragsbuch, volume 25, page 58, with the meeting dates handwritten as 5. 8. 73 – 10. 8. 73, the title {3,4}-Transpositions in finite groups, and below it Fischer’s own signature. And the abstract he wrote is itself a question:

Is there a finite simple group $G$ satisfying …

What follows is the prime factorization of the 33-digit number above. This page makes the word “prediction” solid: at that moment, this group really was still only a question with a precise order.

And right around the time he computed the Baby Monster, Fischer and another person realized almost at the same time: this colossus was very likely only a cross-section of some larger group. That “larger group” is the Monster: at this moment it was predicted for the first time.

§2 · The atlas and the report written in advance

The moment of prediction. At the end of 1973, Fischer noticed that his huge group might be embedded in a larger group; almost at the same time, Robert Griess (1945–) in Michigan, in the United States, independently arrived at the same idea. Both were convinced: there should exist a larger simple group, with the Baby Monster inside it. As for how big it actually was, nobody knew. This is the moment the Monster was “predicted”: people first became convinced that it should exist and how big it should be, while nobody had yet seen it.

In the next few years, what mathematicians did was give this monster, which had not yet shown itself, a medical examination. And the reason this could be done rested on a big book that was being compiled, and a few iron laws about “the fingerprint of a group”.

That big book is called the ATLAS of Finite Groups. The idea arose around 1970 in Cambridge: Conway discovered a huge new group in 1969 (coming from a structure called the Leech lattice, told in EP6), and then spent more than a year working out its properties and computing its fingerprint table; this gave him an idea: simply compile a handbook listing all the interesting properties of all the interesting groups. The book was finally compiled by five people: Conway, Curtis (Robert Curtis), Norton (Simon Norton), Parker (Richard Parker), Wilson (Robert Wilson); it was published by Oxford University Press in 1985, having taken about fifteen years altogether. (An Easter egg only group theorists will smile at: these five surnames all happen to have six letters.)

John Thompson (1932–): his “order formula” can work back from the neighbor circle of an involution to the order of the whole group, and it was the key tool for weighing the Monster. (Photo by Renate Schmid, 2007, Oberwolfach Photo Collection, CC BY-SA 2.0 de.)

The order of the Monster was computed. Weighing such a group that has not yet been built uses a clever tool, the Thompson order formula (invented by Thompson, the group theorist of EP5 and EP6): it can work back from the “centralizer” of an involution (all the elements that commute with it) to the order of the whole group.

This step is worth pausing on. Look closely at what it takes in and what it puts out: what it takes in is only the “neighbor circle” of an involution, the subgroup formed by those elements that commute with it under multiplication; what it puts out is the order of the whole group, an exact integer. It needs only one premise: the involutions of this group fall into exactly two classes; and the Monster has exactly two classes. So what has to be taken in is the neighbor circle of each of these two classes, and no other input is needed beyond that.

The numbers on the Monster’s side make the gap clear. Its involutions fall into two classes; the neighbor circle of the second class (written 2B) is exactly $2^{1+24}.\mathrm{Co}_1$: the Conway group from the previous episode is back. Take $|\mathrm{Co}_0|=8{,}315{,}553{,}613{,}086{,}720{,}000$, already computed in EP6 ($\mathrm{Co}_1$ is the half of it left after quotienting out $\{\pm I\}$), and multiply by the $2^{25}$ of the outer shell; the size of this neighbor circle is

$|C_M(2\mathrm{B})|=2^{25}\cdot\frac{|\mathrm{Co}_0|}{2}=139{,}511{,}839{,}126{,}336{,}328{,}171{,}520{,}000,$

a 27-digit number. And the Monster itself is a 54-digit number: exactly twice as many digits. That is to say, the size of this neighbor circle is of the same order of magnitude as the square root of the whole Monster (in fact about one sixth of the square root); compared with each other, it is smaller by a factor of about $5.8\times10^{27}$. It is just these two small local pieces that fixed that 54-digit number without a single digit wrong.

The intuition for why this road works is that involutions are used in pairs. Take any two involutions; what they generate is always a dihedral group, the symmetry group of some regular polygon, whose number of sides is the order of their product (the lemma in Appendix A proves this, and it holds for any group). So “how big is the whole group” can be rewritten as “how many pairs of such involutions can be formed, and which dihedral block each pair falls into”. Thompson’s formula counts this out to the end; once it is counted, the sizes of the two neighbor circles are multiplied and added, and what is left is exactly $|G|$. (For the shape of the formula see the end of Appendix B; for the complete proof see Appendix D: it is the cheapest proof in the whole article.)

Put it back on the main line of local analysis from EP5 §3: Brauer–Fowler 1955 (fully proved in EP5 Appendix I) says “fix the neighbor circle of an involution, and only finitely many finite simple groups fit it”, collapsing the infinite into the finite; Thompson collapsed this sentence one notch further: not finitely many, but one exact number. The same knife, sharpened a little more.

John Conway (1937–2020): he named Fischer’s new group the “Baby Monster”, and computed the order of the Monster through the night, writing to tell Fischer at daybreak. (Photo by Thane Plambeck at the 2005 Banff conference, CC BY 2.0.)

Back to 1973. Fischer first pinned down an upper bound with two cross-sections; Conway then ran an old programmable calculator all night, got a concrete number at daybreak, tried it, and it fit. He immediately wrote to Fischer (in translation): “Dear Bernd, the order of the Monster is … but I think you already know.” In January 1974, in Oberwolfach, the order of the Monster was written down for the first time, and at that time it did not yet have a name:

$|M| = 2^{46}\cdot3^{20}\cdot5^{9}\cdot7^{6}\cdot11^{2}\cdot13^{3}\cdot17\cdot19\cdot23\cdot29\cdot31\cdot41\cdot47\cdot59\cdot71 \approx 8\times10^{53}.$

(This number has 54 digits. Note that $11$ is squared and $13$ is cubed, not ordinary first powers: this point will come back later.)

Next comes that fingerprint table. Every finite group has a character table: sort the elements of the group by “same type” into a number of classes, then list all the “representations that cannot be broken down” of the group one by one, fill each cell with a number, and you get a square matrix: it is the group’s most condensed identity card. The Monster’s table has 194 rows and 194 columns, and it was computed by Fischer and Donald Livingstone, with the program written by Michael Thorne, in the 1970s: Fischer computed for a long time in Birmingham from about 1974, and it was finalized in 1979 (the years differ slightly across three sources). Remember this time: whichever year you take, the Monster had at that point still not been proved to exist. (What a character is, how this table is read, and on what grounds “computing a group’s table without having seen the group” is possible, see Appendix B: there a small $3\times3$ table computes each of these for you.)

Credit for one thing has to be sorted out here, because it is easily mixed up: the order was fixed by Thompson’s formula, together with the combined efforts of Conway, Harada, Thompson and others; the character table was computed by Fischer, Livingstone and Thorne; as for the proof of this monster’s uniqueness, and the final proof of its existence, those were the business of other groups of people again (uniqueness was reduced by Thompson in 1979 and later completed by Griess and others; existence, that is, building it, is Griess’s work in the next section). These links each belong where they belong; they were not all done by one person.

The most important number in the report: 196883. Look at the first column of that character table (it lists the dimension of each “representation that cannot be broken down”): at the top is always $1$, and the next, second-smallest dimension, for the Monster, is 196883. What it means is: the smallest space in which the Monster can stretch its limbs (over the complex numbers) is 196883-dimensional; with one dimension fewer, it cannot fit. Even more worth mentioning,

$196883 = 47\times59\times71,$

which is exactly the product of the three largest prime factors of the Monster’s order. First, how “the three largest” are determined: look back at the factorization of $|M|$ at the start of §2; the prime factors, arranged from small to large, run at the end exactly $\cdots 31\cdot 41\cdot 47\cdot 59\cdot 71$; these last three are its largest prime factors, with $71$ the largest of them.

As for why the Monster’s prime factors are exactly this set, stopping at $71$, with nothing larger: that is a much deeper question than this, and the answer is hidden in a secret bridge between the Monster and another piece of mathematics. We leave it for the last episode of this series, and then go and see what is behind that door.

Something as large as $8\times10^{53}$: the smallest place it lives in is a 196883-dimensional space.

Figure · The Monster, as large as $8\times10^{53}$, has as the smallest space in which it can stretch its limbs only 196883 dimensions ($=47\times59\times71$): a colossal thing that fits exactly into its smallest room. (Made by this channel)

At this point the medical report is complete: size, fingerprint table, smallest home, all written, while the monster itself has not yet shown its face. In the next section, someone is going to really build it.

§3 · Building the monster by hand

By the end of the 1970s the situation was a little peculiar: all the technical data on the Monster (the order, the 194×194 character table, 196883, and even the 196884 about to come on stage below) was ready, but nobody had proved that it really existed. Conway later recalled his feeling at the time, saying that he regarded building the Monster as “a practical impossibility, or maybe an impractical possibility”.

Robert Griess (1945–): in 1980, without using a single computer, he built the Monster by hand. (Photo by Renate Schmid, 2005, Oberwolfach Photo Collection, CC BY-SA 2.0 de.)

The one who broke this deadlock was Robert Griess. He was one of the first (around 1973) to produce evidence that the Monster exists. And he had in hand an extra clue passed to him by someone else: Norton deduced from that character table that the Monster must preserve a 196884-dimensional algebraic structure; that is, there is a “multiplication” on that 196884-dimensional space, and the Monster is exactly the full set of symmetries that do not scramble this multiplication.

This road sounds even harder than the original one: the thing to be built is bigger. But it is in fact easier: that group has $8\times10^{53}$ elements, and you cannot write them down one by one; an algebra does not need to be written out one by one; giving a multiplication rule is enough. So the route was clear: first build this 196884-dimensional algebra, then prove that “the full set of symmetries preserving it” is exactly a new simple group, precisely the monster that had been predicted and whose medical report had already been written. On January 14, 1980, Griess announced in Ann Arbor: he had done it. Two years later the complete paper (titled The Friendly Giant, Invent. Math. 69) was printed, one hundred and two pages.

The most remarkable thing about this is that he did it by hand, without using a single computer. As a comparison: someone at the time did the arithmetic, and just doing one such matrix multiplication would take up a machine for seventeen or eighteen days: the matrices have nearly two hundred thousand rows and two hundred thousand columns, and by the most naive algorithm $8\times10^{15}$ multiplications are needed; even at five billion per second, it would take seventeen or eighteen days. One multiplication, seventeen or eighteen days. And Griess completed the whole thing with paper and pen. Tits later said (in translation): “I simplified Griess’s construction slightly, but what he did is great work. It was done without a computer, which is a remarkable feat.” Griess himself recalled that he married in June 1979 and from October computed day and night without a break, “taking only half a day off for Thanksgiving and only one day for Christmas” (in translation).

A notation has to be made clear here, or experts can easily find fault: 196883 and 196884 differ by only $1$, but refer to two things.

• 196883 is the dimension of the Monster’s representation: the smallest space in which it can stretch its limbs.
• 196884 is the dimension of that algebra. The usage current today calls this algebra the Griess algebra; the one dimension it has beyond the representation space is the direction of the algebra’s identity element (the "$1$" that leaves anything unchanged when it multiplies it): $196884 = 196883 + 1$.

(A point of historical precision: the algebra constructed in Griess’s original 1982 paper is in fact 196883-dimensional and contains no identity element; “adding a dimension for the identity to make 196884” is a cleaner later restatement. So one cannot say “Griess’s original paper constructed a 196884-dimensional algebra”. In this episode we use the current modern 196884 version, which is both correct and avoids this pitfall. For how the algebra is divided into blocks internally and where the multiplication comes from, see Appendix C.)

And the relation between the Monster and this algebra can be wrapped up with one equals sign. Here $\mathrm{Aut}$ (automorphism group) is shorthand for “the full set of symmetries of some object”: all the transformations that keep this object’s structure exactly as it is, without the slightest disturbance, put together form a group. It runs completely parallel to the chain of equals signs in EP6:

$M_{24}=\mathrm{Aut}\,S(5,8,24),\qquad \mathrm{Co}_0=\mathrm{Aut}(\Lambda_{24}),\qquad M=\mathrm{Aut}(\text{Griess algebra}).$

The three objects in parentheses have all been seen with our own eyes earlier: $S(5,8,24)$ is the Steiner design of EP6 ($24$ points, $8$ points per block, any $5$ points lie in exactly one block), and $M_{24}$ is the full set of rearrangements that keep its block structure exactly as it is; $\Lambda_{24}$ is the Leech lattice of EP6 (the densest packing in $24$ dimensions, with kissing number $196560$), and $\mathrm{Co}_0$ is the full set of orthogonal transformations preserving it; and the Griess algebra is exactly the $196884$-dimensional multiplication just built by hand in §3 of this episode, with $M$ the full set of symmetries that do not scramble this multiplication.

All along the way, the sporadic groups have each been “the full set of symmetries of some exceptionally intricate object”. The Monster is the largest one on this line.

The largest one in the box has at this point shown itself. But the extra "$1$" on it ($196884 = 196883 + 1$) looks unimportant at the moment: its weight will only show in the last episode.

§4 · Twenty-six, and a proof that cannot be read to the end

With the Monster built, a very natural question pops up: is the box empty? Do the sporadic groups end here?

The count had in fact long since closed. The Monster and the several new subgroups it brought along pushed the total number of sporadic groups to 25; in 1975, Janko found evidence for the 26th (written $J_4$), and after that this number never changed again: 26 sporadic simple groups in all. In the previous episode we described how they split into families: of the 26, 20 are subgroups or quotient groups of the Monster, and the remaining 6 cannot be swallowed even by the Monster. And the one who named these two batches was exactly the person who built the Monster by hand: Griess, in The Friendly Giant, fixed both names in one sentence: the 20 that get in are called the “Happy Family”; those that do not, we already named in the previous episode: pariahs. His original words are:

“The sporadic groups which are involved in the Friendly Giant constitute the Happy Family and those which are not are called the Pariahs.” —Griess, The Friendly Giant, Invent. Math. 69 (1982), p. 3

But one detail is worth pausing on: in 1982 Griess could name only five pariahs ($\mathrm{Ly}$, $J_3$, $J_4$, $\mathrm{O'N}$, $\mathrm{Ru}$); for the sixth, $J_1$, he could not give an answer at the time. On the same page he wrote: “The membership of every sporadic group in one of those two categories is settled, except for $J_1$”, and added a very vivid remark: the order of $J_1$ is “only” 175,560, and “one might easily imagine a copy of $J_1$ floating as a tiny speck within $F_1$”.

In other words: the person who built the Monster by hand could not say for sure whether a small group of order one hundred seventy thousand was hidden inside the thing he had built. This suspense was only resolved in 1986 by Wilson (Robert Wilson), and the title of the paper is itself the question: “Is $J_1$ a subgroup of the Monster?” (Is $J_1$ a Subgroup of the Monster?, Bull. London Math. Soc. 18(4): 349–350, 1986, doi:10.1112/blms/18.4.349). The answer is no, and so the pariahs made up the full 6, becoming the number we write today.¹

So, “has the classification of finite simple groups been completely finished?” The answer to this question brings out a theme that recurs in this series: a proof so long that nobody can read it to the end alone.

In 1983, Gorenstein (Daniel Gorenstein, 1923–1992) announced: the classification of finite simple groups has been fully completed. But this announcement came too early: he was misled by one proof into thinking that one class of cases, called “quasithin groups”, had been sealed, when in fact a gap remained there. Those who really filled this gap were Michael Aschbacher and Stephen Smith, and by then it was 2004: a proof in two large volumes totaling about 1221 pages. Only in that year was the classification of finite simple groups generally accepted as truly complete.

From Gorenstein’s proposal of the program in 1972 to the filling of the last piece in 2004, this classification project, which he himself called the “Thirty Years’ War”, spanned just about thirty years in all: that is exactly the theme of the EP5 episode. The appearance of the Monster (1980) falls in the middle of this war: it is the largest and also the most daunting item on this list that cannot be read to the end.

Epilogue · The box still has a crack

The Monster has shown itself. It is large beyond intuition ($8\times10^{53}$), yet it is an object that is finite, concrete, and can really be built by a pair of hands. It lives in 196883 dimensions, and its data has been checked again and again: and the people who checked it say themselves that the question of reliability has not gone away (see §2). It looks as if the story could wrap up here.

But one number will not keep still. The dimension of the Monster’s smallest representation is 196883, and the dimension of that algebra is 196884. And in a corner seemingly unrelated to group theory and to symmetry (a classical object in number theory called the $j$-function), the first interesting coefficient of its expansion happens to be exactly 196884.

How can a number that describes the most space-saving form of the monster appear, without a digit’s difference, in a number-theory formula written down more than a hundred years earlier? Is it a coincidence, or is there one more thing hidden in the deepest part of the box?

The box still has one last crack. In the last episode we go and see what is behind that crack.

Appendix A (intuition + structure) · 3-transposition groups and mirror networks

Figure A · Reflecting in two mirrors one after the other composes to a rotation, by twice the angle between them; so “the order of the product is at most 3” is the same as “the angle may only be coincidence, $90^\circ$, $60^\circ$”, and $45^\circ$ is out. In the middle is the mirror network formed by the six transpositions of $S_4$: the three solid lines are the only three pairs of disjoint transpositions (product of order 2), and the other twelve pairs all share a point (product of order 3). On the right are the three new sporadic groups that Fischer’s theorem closes on. The sign in the corner is the iron rule of notation: Fischer’s $\mathrm{Fi}$ is not Mathieu’s $M$. (Made by this channel · math infographic)

What this section does. Section §1 of the main text says that Fischer (Bernd Fischer), with the seemingly plain question of “groups generated by transpositions”, dug out three brand-new sporadic groups. Here we make “3-transposition group” precise, use the $S_n$ you know best as the instance, prove the statement “two mirrors compose to a rotation” that we have kept using as intuition, and then give the statement and intuition of Fischer’s theorem. The ceiling of this episode takes effect here: give the structure and examples; the complete classification proof is cited, not rebuilt.

Concept explained here: what is a 3-transposition group. A group $G$ generated by some conjugacy class $D$ (“conjugacy class” = a whole class of elements of the same type; EP4 Appendix 1 §2 gave the definition: the conjugacy class of $g$ = all $hgh^{-1}$), where $D$ consists entirely of involutions (an involution returns to the identity after multiplying by itself once, $d^2=e$: the entrance key drawn out of the odd order theorem in EP5 §3), and for any $d,e\in D$, the order of the product $de$ may only be $1$, $2$ or $3$. Such a $(G,D)$ is called a 3-transposition group. The three numbers each cover one case: order $1\iff d=e$; order $2\iff$ the two involutions commute; order $3\iff$ together they give rise to a small, triangle-like symmetry (we will see it with our own eyes in $S_n$ shortly).

Mirror intuition. An involution is like a mirror: reflecting twice = not reflecting. Reflecting in two mirrors in succession composes to a rotation, with angle = twice the angle between the two mirrors; so “the order of $rs$ is $n$” translated into geometry is “the angle = $180^\circ/n$”. “Product order $\le 3$” is therefore a rule for placing mirrors: any two mirrors either coincide, or meet at $90^\circ$, or meet at $60^\circ$, and no other angle is allowed. The whole group becomes a huge mirror network; what Fischer asked was: how many kinds of such networks can there be in all?

Below we set the two pillars of this intuition firmly in place one by one: first prove “two involutions always generate a dihedral group” (pure group theory, no mirrors needed), then prove “composing two mirrors = doubling the angle of rotation” (pure plane geometry), and finally put them together and see why that mirror-placing rule leaves exactly three angles.

Pillar one: any two involutions always generate a dihedral group (full proof).

Lemma. In a group $G$, take any two involutions $d,e$ (that is, $d^2=e^2=1$). Write $\rho=de$, $m=\mathrm{ord}(\rho)$. Then $d\rho d^{-1}=\rho^{-1}$, and $\langle d,e\rangle$ is the dihedral group of order $2m$, $D_m$ (the symmetry group of the regular $m$-gon).

Proof. Since $d^{-1}=d$, $e^{-1}=e$, first compute the inverse: $\rho^{-1}=(de)^{-1}=e^{-1}d^{-1}=ed$. Then compute the conjugate: $d\rho d^{-1}=d(de)d=(dd)(ed)=ed=\rho^{-1},$ that is, $d$ turns $\rho$ into its inverse. Also $d\rho=d(de)=(dd)e=e$, so $e$ is generated by $d,\rho$, and $\langle d,e\rangle=\langle d,\rho\rangle$. The cyclic subgroup $\langle\rho\rangle$ (of order $m$) is normalized by $d$ (the equation above), so $\langle d,\rho\rangle=\langle\rho\rangle\cup d\langle\rho\rangle$, of order at most $2m$. When $m\ge3$: if $d\in\langle\rho\rangle$, then $\langle\rho\rangle$ is commutative, and the equation above gives $\rho=\rho^{-1}$, $m\le2$, a contradiction; so $d\notin\langle\rho\rangle$, and the order is exactly $2m$. And $d^2=\rho^m=1$, $d\rho d^{-1}=\rho^{-1}$ are precisely the defining relations of the dihedral group $D_m$. $\blacksquare$

($m=1$: $d=e$, and $\langle d\rangle$ has order $2$; $m=2$: $d,e$ commute, giving the Klein four-group: in EP5 §3, the involution centralizer in $A_5$ of order $4$ is exactly this. The two degenerate cases agree with the usual conventions for $D_1,D_2$.)

This lemma is universal: “the order of $de$ is $m$” and ”$d,e$ generate the symmetry group of the regular $m$-gon” are the same sentence. So “mirror” is not a figure of speech: the structure spanned by two involutions is, from the start, the symmetry group of a regular polygon.

Pillar two: composing two mirrors = a rotation, with angle = twice the angle between them (full proof).

For a line in the plane through the origin making with the $x$-axis an angle $\theta$, the matrix of reflection in it is

$R_\theta=\begin{pmatrix}\cos2\theta&\sin2\theta\\ \sin2\theta&-\cos2\theta\end{pmatrix}.$

(Where it comes from: the unit vector making with the $x$-axis an angle $\varphi$, $(\cos\varphi,\sin\varphi)$, has a mirror image in this line that makes with the $x$-axis an angle $2\theta-\varphi$; read $\cos(2\theta-\varphi)=\cos2\theta\cos\varphi+\sin2\theta\sin\varphi$, $\sin(2\theta-\varphi)=\sin2\theta\cos\varphi-\cos2\theta\sin\varphi$ as the action of a matrix and you get it. You can check it on the spot: $\theta=0$ gives $\mathrm{diag}(1,-1)$, which is exactly folding about the $x$-axis; $\theta=45^\circ$ gives $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$, which is exactly swapping the two coordinates along $y=x$.)

Take two mirrors $R_\alpha$, $R_\beta$, with angle $\phi=\beta-\alpha$ between the two mirror lines. Multiply directly:

The four cells are, one by one, the angle-sum formulas: top left $\cos2\beta\cos2\alpha+\sin2\beta\sin2\alpha=\cos2(\beta-\alpha)$; top right $\cos2\beta\sin2\alpha-\sin2\beta\cos2\alpha=-\sin2(\beta-\alpha)$; bottom left $\sin2\beta\cos2\alpha-\cos2\beta\sin2\alpha=\sin2(\beta-\alpha)$; bottom right $\sin2\beta\sin2\alpha+\cos2\beta\cos2\alpha=\cos2(\beta-\alpha)$. The right-hand side is exactly the rotation matrix for angle $2\phi$. $\blacksquare$

Corollary: that mirror-placing rule leaves only three angles. A rotation by angle $2\phi$ has order $n$ if and only if $2\phi$ is $360^\circ$ times $1/n$ (in lowest terms), that is, the angle between the mirror lines is $\phi=180^\circ/n$. Substituting one by one:

Counterexample ($45^\circ$). Two mirrors at $45^\circ$ compose to a $90^\circ$ rotation, of order $4$, beyond $\{1,2,3\}$. By the lemma of pillar one, these two involutions generate $D_4$, of order $8$, the symmetry group of the square. So: the symmetry network of the square as a whole cannot get onto Fischer’s list; the largest regular polygon that can get in is the equilateral triangle ($D_3\cong S_3$, $60^\circ$). The mirror-placing rule “only coincidence / $90^\circ$ / $60^\circ$” is at this point no longer a convention, but all the values of $180^\circ/n$ for $n\le3$.

Honest boundary (an important point). The two pillars above have different strengths: the lemma is universal: in any group, any two involutions generate $D_m$, without exception. The two-mirror theorem is plane geometry: it speaks about that geometric realization of $D_m$. In a general 3-transposition group, $D$ is a conjugacy class of involutions, not necessarily actual mirrors in some plane (for example, for those families of classical groups over $\mathbb{F}_2$, “mirror” has to be understood in the geometry over a finite field; cited, not rebuilt). On the $S_n$ side, though, it holds literally: the transposition $(i\;j)$ acting on $\mathbb{R}^n$ is exactly the orthogonal reflection in the hyperplane $x_i=x_j$ (it fixes that hyperplane and reverses the sign of $e_i-e_j$); in the instance below, the two methods of computation will match on the same answer.

Instance: $S_n$ itself is a 3-transposition group. Take $D=$ all transpositions $(i\;j)$ (swap two points, leave the rest unmoved; cycle notation was just learned in EP6 Appendix B). Taking any two transpositions, there are exhaustively only three possible relative positions:

• Equal: $(1\;2)(1\;2)=e$, order $1$.
• Disjoint: $(1\;2)(3\;4)$: each moves its own points without interfering, they commute, and the product has order $2$.
• Sharing one point: $(1\;2)(1\;3)=(1\;3\;2)$, a $3$-cycle, order $3$.

There is no fourth case, so the product order is always $\in\{1,2,3\}$: $S_n$ is a 3-transposition group (checked by enumerating pair by pair over the transpositions of $S_6$, all $15$ of them, $15\times15$ without exception).

The two roads match. Recompute the three lines above with mirrors: the mirror of the transposition $(i\;j)$ is the hyperplane $x_i=x_j$, whose normal vector (the “root”) is $e_i-e_j$; the angle between two mirrors = the angle between the two roots. For the disjoint $(1\;2),(3\;4)$, the roots $e_1-e_2$ and $e_3-e_4$ are orthogonal, the angle is $90^\circ$ → rotation angle $180^\circ$ → order $2$. For $(1\;2),(1\;3)$ sharing a point, the angle between the roots $e_1-e_2$ and $e_1-e_3$ is given by $\cos\vartheta=\frac{(e_1-e_2)\cdot(e_1-e_3)}{|e_1-e_2|\,|e_1-e_3|}=\frac{1}{\sqrt2\cdot\sqrt2}=\frac12$ as $\vartheta=60^\circ$ → rotation angle $120^\circ$ → order $3$. Line by line in agreement with the orders obtained by counting cycles directly (both methods went into the check). $S_3$ is the smallest nontrivial mirror network: three mirrors, pairwise at $60^\circ$.

Fischer’s theorem (1971, cited at the level of statement). Under technical hypotheses of the kind “no nontrivial solvable normal subgroup”, the list of finite 3-transposition groups is abnormally short: the symmetric groups $S_n$; three families of classical groups over $\mathbb{F}_2$ (symplectic groups $\mathrm{Sp}_{2n}(2)$, unitary groups $\mathrm{PSU}_n(2)$, orthogonal groups $\mathrm{O}^{\pm}_{2n}(2)$, all on the “Lie-type families” list of EP5 §1, of which the unitary groups were constructed explicitly in EP5 Appendix D); one family of orthogonal-type groups over $\mathbb{F}_3$; and exactly three new groups that nobody had seen before: $\mathrm{Fi}_{22}$, $\mathrm{Fi}_{23}$, $\mathrm{Fi}_{24}$. The iron rule of notation struck once more (the mine planted in the EP6 epilogue): Fischer’s $\mathrm{Fi}$ and Mathieu’s $M$ are two different families: $|\mathrm{Fi}_{22}|=2^{17}\cdot3^{9}\cdot5^{2}\cdot7\cdot11\cdot13=64{,}561{,}751{,}654{,}400\approx6.5\times10^{13}$, about one hundred fifty million times larger than $|M_{22}|=443{,}520$. $|\mathrm{Fi}_{23}|=2^{18}\cdot3^{13}\cdot5^{2}\cdot7\cdot11\cdot13\cdot17\cdot23=4{,}089{,}470{,}473{,}293{,}004{,}800$. $\mathrm{Fi}_{24}$ itself is not simple: it contains a normal subgroup taking up exactly half of it (the bookkeeping of Lagrange’s theorem in EP2 group theory mini-lesson ④: index $2$), and it is that half that is the sporadic simple group, written $\mathrm{Fi}_{24}'$ (the prime is the mark for “take that half”), $|\mathrm{Fi}_{24}'|=2^{21}\cdot3^{16}\cdot5^{2}\cdot7^{3}\cdot11\cdot13\cdot17\cdot23\cdot29=1{,}255{,}205{,}709{,}190{,}661{,}721{,}292{,}800$. All three orders checked against the ATLAS (the products have been computed). The complete proof of the theorem belongs to Fischer’s original paper “Finite groups generated by 3-transpositions. I” (Invent. Math. 13 (1971) 232–246), cited, not rebuilt.

From $3$ to $\{3,4\}$: the ramp to the Monster (details of the historical line belong to main text §1–§2). Loosen the mirror-placing rule by one notch, allowing product orders up to $4$ (so-called $\{3,4\}$-transposition groups; by the table above, this is exactly letting $45^\circ$ in), and push the same mirror-network philosophy further: along this ramp Fischer first touched the Baby Monster $B$; and the prediction of the Monster $M$ (around 1973, proposed independently by Fischer and Griess (Robert Griess)) was exactly “there exists a simple group in which the centralizer of some involution contains a double cover of $B$” (“double cover” = a shelled version exactly twice as large as the original group, which goes back to it after quotienting out the center $\{\pm\}$: in EP6, $\mathrm{Co}_0$ stands in exactly this relation to $\mathrm{Co}_1$). The involution centralizer, that knife of local analysis from EP5 §3, is drawn from its sheath for the third time.

Symbolic check: the transpositions of $S_6$, all $15$, enumerated pair by pair ($225$ pairs), product order always $\in\{1,2,3\}$, and “equal $\to1$ / disjoint $\to2$ / sharing a point $\to3$” holds pair by pair; on the same $225$ pairs $|\langle d,e\rangle|=2\,\mathrm{ord}(de)$ holds pair by pair (dihedral lemma); $(1\;2)(1\;3)=(1\;3\;2)$ has order $3$; the two-mirror theorem fully proved as a symbolic identity ($R_\beta R_\alpha$ equal cell by cell to the rotation matrix for angle $2(\beta-\alpha)$, not a numerical approximation), and $R_\theta^2=I$, $\det R_\theta=-1$; cross-check of the mirror realization: root angles $90^\circ/60^\circ$ agree line by line with permutation product orders $2/3$; the product of two mirrors at $45^\circ$ has order $=4$ and generates $D_4$ (order $8$); the products of prime factors for the three Fischer group orders agree digit by digit with the decimal integers (against the ATLAS); $|\mathrm{Fi}_{22}|/|M_{22}|=145{,}567{,}477.5\ldots\approx1.5\times10^{8}$. The list in Fischer’s theorem itself = citation (the 1971 original), not part of the check.

Appendix B (intuition + structure) · The character table and 196883

Figure B · The character table of the Monster: the number of irreducible complex representations = the number of conjugacy classes, so this table must be square: 194 across and 194 down. The first column lists the dimension of each representation; at the top is $1$, and immediately after it, the second, is $196883=47\cdot59\cdot71$ (the three largest prime factors in the Monster’s order). The lower left corner uses $S_3$‘s small $3\times3$ table to lay out the same two rules and compute them: row orthogonality was already proved in EP4 Appendix 1, column orthogonality is the one added in this episode, and $1^2+1^2+2^2=6$ is just its special case on the first column. The timeline at the top is the crux of this episode: 1973 prediction, 1979 table finished, and only in 1980 did someone build it. (Made by this channel · math infographic)

What this section does. Section §2 of the main text says the Monster’s “medical report” existed before the Monster itself: the $194\times194$ character table had been computed by 1979, while the group would only be built in 1980. Here we make clear: what a character is (the four words planted in EP4 become regular here), how this table is read, which cell of the table $196883$ stands in, and on what grounds “computing a group’s table without having seen the group” is possible. This section is the engine of this episode’s core claim, so it is one notch harder than elsewhere: the theorem that makes the whole thing hold (the table must be square), together with the column orthogonality relation, is proved all the way here (EP4 Appendix 1 only proved the row half). The Monster’s table itself is cited from the ATLAS, not recomputed.

Concepts explained here (calling back to EP4). These four words were given in full, with definitions and proofs, in EP4 Appendix 1 (“Foundations of representation theory: complete reducibility, characters, Schur orthogonality”); here we wake them up and reuse them: representation (faithfully translating group elements into matrices, $g\mapsto\rho(g)$, with the multiplication structure preserved), dimension (the size $n$ of the matrices), irreducible (the smallest blocks that cannot be broken down: any representation can be broken into a platter of irreducible blocks, the complete reducibility of EP4 Appendix 1 §1), character ($\chi(g)=\operatorname{tr}\rho(g)$, the sum of the diagonal entries of the matrix, the “fingerprint” of a representation). A key convenience: the trace does not move under change of basis, so elements in the same conjugacy class have the same fingerprint: a character only needs one number recorded per class (such a function is called a class function: a function taking a constant value on each conjugacy class).

Arranging them in a square matrix gives the character table: rows = irreducible representations, columns = conjugacy classes, and cell $(i,j)$ = the trace of the $i$-th irreducible representation on the $j$-th class. The first column (the column of the identity element) is exactly the dimensions of the representations, because $\chi(e)=\operatorname{tr}I_n=n$.

And “number of rows = number of columns” (the number of irreducible complex representations is exactly equal to the number of conjugacy classes) is not a coincidence of notation but a theorem. (The word “complex” cannot be dropped: over the real numbers, or when the characteristic $p$ divides $|G|$, this equality no longer holds. Wherever the text below and the figures simplify, complex representations are meant throughout.) It is the foundation of this episode’s statement “computing a group’s table without having seen the group”, and we prove it below.

Waking up: the $S_3$ table of EP4 Appendix 1 §3. $S_3$ (the symmetry group of the equilateral triangle dissected throughout the group theory mini-lessons of EP2) has $3$ conjugacy classes: $\{e\}$, the three transpositions, the two three-cycles, of sizes $1,3,2$ respectively; it has exactly $3$ irreducible representations, whose characters are written $\chi_1$ (trivial), $\chi_{\mathrm{sgn}}$ (sign), $\chi_{\mathrm{std}}$ (standard two-dimensional). The whole table, together with the term-by-term check of row orthogonality, was already done in EP4 Appendix 1 §3; we do not re-derive it here, but just take the three rows of values as raw material for what follows:

$\chi_1=(1,\;1,\;1),\qquad \chi_{\mathrm{sgn}}=(1,\;-1,\;1),\qquad \chi_{\mathrm{std}}=(2,\;0,\;-1),$

(The three columns correspond in order to the classes $\{e\}$, transpositions, three-cycles, with class sizes $1,3,2$.) What EP4 checked on this table was the orthogonality of row against row; what this section is to prove is that column against column is orthogonal too: the same table, turned sideways and used once more.

Theorem 1: the table must be square (full proof).

Theorem. For a finite group $G$, the number of irreducible complex representations (up to isomorphism), $r$, is exactly equal to the number of conjugacy classes of $G$, $k$.

Proof. Write $\mathcal{C}$ = the complex vector space formed by all class functions on $G$, with the inner product carried over from EP4 Appendix 1 §3, $\langle\alpha,\beta\rangle=\frac{1}{|G|}\sum_{g}\alpha(g)\overline{\beta(g)}$. In three steps.

Step 1: $\dim\mathcal{C}=k$. A class function takes one constant value on each conjugacy class, and these constant values can be specified arbitrarily; so take the indicator functions of the classes $1_{K_1},\dots,1_{K_k}$ (on $K_j$ taking $1$, elsewhere $0$); they are clearly linearly independent and span $\mathcal{C}$. Hence $\dim\mathcal{C}=k$.

Step 2: $r\le k$. EP4 Appendix 1 §3 already proved the first orthogonality relation $\langle\chi_i,\chi_j\rangle=\delta_{ij}$: the irreducible characters are a set of orthonormal vectors in $\mathcal{C}$, and therefore linearly independent. An orthogonal set in a $k$-dimensional space has no more than $k$ members, so $r\le k$.

Step 3: they also span $\mathcal{C}$, so $r=k$. It suffices to prove: if $\alpha\in\mathcal{C}$ is orthogonal to every $\chi_i$, then $\alpha=0$ (orthogonal complement is zero $\iff$ spanning). Suppose $\langle\alpha,\chi_i\rangle=0$ for all $i$. For each irreducible representation $(\rho_i,V_i)$, construct the operator

$T_i=\sum_{g\in G}\overline{\alpha(g)}\,\rho_i(g).$

It is an intertwining operator: for any $h\in G$,

where the change of variable in the middle step is a bijection from $G$ to itself, and the last step is exactly where we use that $\alpha$ is a class function ($\alpha(h^{-1}g'h)=\alpha(g')$). Hence $\rho_i(h)T_i=T_i\rho_i(h)$.

By Schur’s lemma (b) (already proved in EP4 Appendix 1 §3), $T_i=\lambda_i\,\mathrm{Id}$. Take the trace to determine $\lambda_i$: $n_i\lambda_i=\operatorname{tr}T_i=\sum_g\overline{\alpha(g)}\,\chi_i(g)=|G|\cdot\frac{1}{|G|}\sum_g\chi_i(g)\overline{\alpha(g)}=|G|\,\langle\chi_i,\alpha\rangle=0,$ so $\lambda_i=0$, that is, $T_i=0$: this holds for every irreducible representation.

Then by complete reducibility (already proved in EP4 Appendix 1 §1): any representation is a direct sum of irreducible representations, and the construction of $T$ proceeds block by block on a direct sum, so for any representation $\rho$ we have $\sum_g\overline{\alpha(g)}\rho(g)=0$. In particular take the regular representation (the one used when counting multiplicities at the end of EP4 Appendix 1 §3: with the group elements as basis $\{e_x\}_{x\in G}$, $\rho_{\mathrm{reg}}(g)e_x=e_{gx}$), and apply this zero operator to the basis vector $e_1$:

$0=\sum_g\overline{\alpha(g)}\,\rho_{\mathrm{reg}}(g)\,e_1=\sum_g\overline{\alpha(g)}\,e_g.$

And $\{e_g\}_{g\in G}$ is a basis, so the coefficients must be zero one by one: $\overline{\alpha(g)}=0$ for all $g$, that is, $\alpha=0$. $\blacksquare$

So the irreducible characters are an orthonormal basis of $\mathcal{C}$, and $r=\dim\mathcal{C}=k$. The character table is a square matrix: as many conjugacy classes as there are, so many irreducible representations there are, not one more, not one fewer.

Theorem 2: the column orthogonality relation (full proof).

Theorem (second orthogonality relation). For $g,h\in G$, $\sum_{i=1}^{k}\chi_i(g)\,\overline{\chi_i(h)}=\begin{cases}|C_G(g)|,&g\sim h\ (\text{same conjugacy class}),\\ 0,&\text{otherwise}.\end{cases}$ where $C_G(g)$ = the subgroup consisting of all group elements that commute with $g$ (the centralizer).

Proof. By Theorem 1, the irreducible representations and the conjugacy classes both number $k$, so the whole table can be packed into a $k\times k$ square matrix $A$:

$A_{ij}=\sqrt{\frac{|K_j|}{|G|}}\;\chi_i(g_j)\qquad(g_j\in K_j\ \text{any one representative}).$

First look at $AA^*$ ($A^*$ = conjugate transpose): $(AA^*)_{ii'}=\sum_{j}A_{ij}\overline{A_{i'j}}=\sum_j\frac{|K_j|}{|G|}\chi_i(g_j)\overline{\chi_{i'}(g_j)}=\frac{1}{|G|}\sum_{g\in G}\chi_i(g)\overline{\chi_{i'}(g)}=\langle\chi_i,\chi_{i'}\rangle=\delta_{ii'}$ (in the middle, “summing by class, multiplying by class size” is restored to “summing over group elements”, because a character takes a constant value on each class). What this step uses is exactly the row orthogonality of EP4 Appendix 1 §3. So

$AA^*=I_k.$

The key step is here: $A$ is a square matrix, so $AA^*=I$ means $A^*=A^{-1}$, and therefore the other side holds too: $A^*A=I_k.$ Write it out: $\delta_{jj'}=(A^*A)_{jj'}=\sum_i\overline{A_{ij}}\,A_{ij'}=\frac{\sqrt{|K_j|\,|K_{j'}|}}{|G|}\sum_{i}\overline{\chi_i(g_j)}\,\chi_i(g_{j'}).$

Read it in two cases:

• $j\ne j'$ (different classes): the right-hand side $=0$, and the coefficient in front is nonzero, so $\sum_i\overline{\chi_i(g_j)}\chi_i(g_{j'})=0$.
• $j=j'$ (same class): $\frac{|K_j|}{|G|}\sum_i|\chi_i(g_j)|^2=1$, that is, $\sum_i|\chi_i(g_j)|^2=\dfrac{|G|}{|K_j|}$. And by the orbit–stabilizer theorem (walked through explicitly in EP5 Appendix H to compute $A_5$‘s $60/15=4$), let $G$ act on itself by conjugation: the orbit of $g_j$ is exactly its conjugacy class $K_j$, and the stabilizer is exactly its centralizer $C_G(g_j)$, so $|G|/|K_j|=|C_G(g_j)|$. $\blacksquare$

Why the square-matrix step cannot be skipped. “Row orthogonality implies column orthogonality” relies entirely on $A$ being a square matrix: for a non-square matrix, $AA^*=I$ does not imply $A^*A=I$. A $1\times2$ example is enough to see it: take $A=\begin{pmatrix}\tfrac{1}{\sqrt2}&\tfrac{1}{\sqrt2}\end{pmatrix}$; then $AA^*=\tfrac12+\tfrac12=1=I_1$ (the “row” is orthonormal), but $A^*A=\begin{pmatrix}\tfrac12&\tfrac12\\\tfrac12&\tfrac12\end{pmatrix}\neq I_2.$ So Theorem 1 (the table is square) is not a polite preliminary to Theorem 2; it is the pile cap on which Theorem 2 stands. This is also where the real weight of the statement ”$194$ conjugacy classes $\Rightarrow$ the table is $194\times194$” lies.

Verifying column orthogonality on $S_3$ (the three rows of values woken up earlier, used sideways). First compute the centralizers: $C_{S_3}(e)=S_3$ of order $6$; $C_{S_3}((1\,2))=\{e,(1\,2)\}$ of order $2$; $C_{S_3}((1\,2\,3))=A_3$ of order $3$. (One can also use $|G|/|K|$ directly: $6/1,\,6/3,\,6/2$; they match.)

• Same column (the sum of squares of the column $=$ the order of the centralizer of that class): $$\underbrace{1^2+1^2+2^2}_{e\ \text{column}}=6=|C(e)|,\qquad \underbrace{1^2+(-1)^2+0^2}_{\text{transposition column}}=2=|C((1\,2))|,\qquad \underbrace{1^2+1^2+(-1)^2}_{\text{three-cycle column}}=3=|C((1\,2\,3))|.$$
• Different columns (pairwise zero): $$\underbrace{1\!\cdot\!1+1\!\cdot\!(-1)+2\!\cdot\!0}_{e\,/\,\text{transposition}}=0,\qquad \underbrace{1\!\cdot\!1+1\!\cdot\!1+2\!\cdot\!(-1)}_{e\,/\,\text{three-cycle}}=0,\qquad \underbrace{1\!\cdot\!1+(-1)\!\cdot\!1+0\!\cdot\!(-1)}_{\text{transposition}\,/\,\text{three-cycle}}=0.$$

Collecting along the way an iron law that previously we only copied down. Look at the first formula above: the sum of squares of the $e$ column $=1^2+1^2+2^2=6=|S_3|$. This is not a coincidence: apply column orthogonality to the column of the identity element ($g=h=e$): $\chi_i(e)=n_i$, and $C_G(e)=G$, so

$\boxed{\;\sum_{i=1}^{k}n_i^2=|C_G(e)|=|G|.\;}$

“Sum of squares of the dimensions = order of the group” is the special case of column orthogonality on the first column. Elsewhere it is often treated as a formula to memorize; here it is proved.

The Monster’s table: a “medical report” that came before the group itself. After the ramp at the end of Appendix A led to the prediction around 1973, what mathematicians had in hand were only assumptions: assume $M$ exists, assume the centralizers of its two classes of involutions look like that (one class tied to the double cover of the Baby Monster, the other $=2^{1+24}.\mathrm{Co}_1$: the $\mathrm{Co}_1$ of EP6 comes back on stage here; $2^{1+24}$ reads as “wearing a binary shell of $1+24$ layers”, and it grows exactly around the Leech lattice). On the strength of these assumptions plus the iron laws of character theory alone, the order was determined by working backward:

$|M| = 2^{46}\cdot3^{20}\cdot5^{9}\cdot7^{6}\cdot11^{2}\cdot13^{3}\cdot17\cdot19\cdot23\cdot29\cdot31\cdot41\cdot47\cdot59\cdot71 \approx 8\times10^{53}$

($54$ decimal digits: $808{,}017{,}424{,}794{,}512{,}875{,}886{,}459{,}904{,}961{,}710{,}757{,}005{,}754{,}368{,}000{,}000{,}000$; the product has been computed). In the 1970s (finalized by 1979; for the discrepancy in years across three sources see §2), Fischer and Livingstone (Donald Livingstone), with the help of a program written by Thorne (Michael Thorne), finished computing the whole $194\times194$ character table: at this point the Monster had still not been proved to exist. One more is missing: the dimension divides the order of the group (short, using the algebraic-integer toolbox). Besides the three constraints above, one more was used again and again in working back to the character table: the dimension $n_i$ of each irreducible representation divides $|G|$. It comes easily: EP4 Appendix 5 §1 already proved that the central character $\omega_i(K_j)=\dfrac{|K_j|\chi_i(g_j)}{n_i}$ is an algebraic integer, and from row orthogonality $\sum_{g}\chi_i(g)\overline{\chi_i(g)}=|G|$, grouping by class and dividing by $n_i$:

$\frac{|G|}{n_i}=\sum_{j}\frac{|K_j|\,\chi_i(g_j)}{n_i}\,\overline{\chi_i(g_j)}=\sum_j\omega_i(K_j)\,\overline{\chi_i(g_j)}.$

The right-hand side is an algebraic integer ($\omega_i$ and $\chi_i$ are both algebraic integers, and the algebraic integers form a ring); the left-hand side is a rational number. And the algebraic integers in $\mathbb{Q}$ are exactly $\mathbb{Z}$, so $|G|/n_i$ is an integer, that is, $n_i\mid|G|$. $\blacksquare$ (Here “the algebraic integers form a ring” and “the algebraic integers in $\mathbb{Q}$ $=\mathbb{Z}$” are number-theoretic background, borrowed from EP4 Appendix 5 §0 and throughout this series cited, not rebuilt: re-proving them belongs to algebraic number theory and is not within this episode’s ceiling.)

On what grounds it is possible. Now we can answer head-on the statement in main text §2, “the medical report came before the thing itself”. With only a few assumptions about involution centralizers in hand, it was enough to force a $194\times194$ table to be almost unique, relying on these constraints just proved, which bite into each other:

1. The table is square (Theorem 1): $194$ conjugacy classes means there must be $194$ irreducible representations, not a single cell more;
2. Rows are orthonormal to each other (EP4 Appendix 1 §3);
3. Columns are orthogonal to each other, and each column’s sum of squares $=$ the order of the centralizer of that class (Theorem 2): the orders of the centralizers are exactly the quantities those assumptions give directly, so the assumptions turn directly into numerical constraints on the table;
4. $\sum n_i^2=|G|$ (the special case of Theorem 2 on the first column) and $n_i\mid|G|$: these pin the sequence of dimensions into a very narrow range.

How tight are these conditions: they do not “help” determine this table; they almost lock it in. What that machine did in 1979 was fill in the $194\times194$ cells one by one under these constraints: what it filled in was the medical report of a monster that at the time nobody had seen, and that had not even been proved to exist.

There is one more machine: computing the order directly. The four constraints above force the table into a very narrow space; but the number $|M|$ itself was given by another tool, the Thompson order formula (main text §2 said what it takes in and what it puts out; here we give its shape): suppose $G$ has exactly two classes of involutions, with representatives $t$, $z$; then

$|G|=|C_G(z)|\cdot a(t)+|C_G(t)|\cdot a(z),$

where $a(x)$ counts those involution pairings $(u,v)$ ($u$ conjugate to $t$, $v$ conjugate to $z$) in which $x$ falls inside $\langle uv\rangle$. By the dihedral lemma of Appendix A, $\langle u,v\rangle$ is a dihedral group and $\langle uv\rangle$ is exactly its rotation half: so what $a(x)$ counts is “in the rotation part of which dihedral blocks $x$ can be seen”. The Monster has exactly “two classes of involutions” (2A, 2B), so the premise fits exactly. And so the whole chain closes: assumptions ⟹ two neighbor circles ⟹ order ⟹ combined with the orthogonality relations ⟹ the whole $194\times194$, all without ever having seen that group. (For the complete proof of the formula see Appendix D: it is cheaper than any passage of this appendix: it needs only that dihedral lemma of Appendix A plus one double count, and does not touch characters.)

Where $196883$ is. Look at the first column of the table (the dimensions of the irreducible representations): at the top is always $1$ (the trivial representation); for the Monster, the second-smallest dimension is exactly $196883$: it is the smallest dimension of a faithful complex representation of $M$ (“faithful” = different group elements correspond to different matrices, explained in EP6 Appendix D; the trivial representation maps everything to $1$, the least faithful). Moreover

$196883 = 47\cdot59\cdot71$

is exactly the product of the three largest in the list of prime factors of $|M|$ (computed). A colossus of $8\times10^{53}$ needs, even for its most frugal matrix home, $196883$ dimensions. (A note on the convention: this is the answer over the complex field $\mathbb{C}$, the ATLAS convention; change the field and the number moves: for example over $\mathbb{F}_2$ the smallest faithful dimension is $196882$; we only touch on this.)

A small piece of the real table. The whole $194\times194$ cannot be printed here (in the ATLAS it takes up several pages, and the cells hold algebraic integers). But the first column (the dimensions of the irreducible representations) can be glanced at. The first ten entries, in ATLAS order:

$1,\quad 196883,\quad 21296876,\quad 842609326,\quad 18538750076,\quad 19360062527,$ $293553734298,\quad 3879214937598,\quad 36173193327999,\quad 125510727015275,\quad\dots$

Three things are worth making clear with these ten numbers.

One: $194$ dimensions, not $194$ “different” dimensions. We proved above that “number of irreducibles = number of conjugacy classes”, so this column has exactly $194$ entries, no more and no fewer; but only $170$ of them are mutually different: some dimensions are shared by two different irreducible representations (the $16$th and $17$th entries, for example, are both $8{,}980{,}616{,}927{,}734{,}375$), and in one place three share one. Same dimension does not mean same representation: they can be told apart in the other columns. This is exactly why the character table has to have $194$ columns: looking only at the first column is not enough.

Two: every dimension divides $|M|$. The statement $n_i\mid|G|$ above can be checked entry by entry here: these ten entries (and indeed all $194$) without exception. This is not a coincidence but a direct consequence of that theorem; back when this table was worked out backward, it was exactly one of the sieves used to filter out candidate dimensions. (A word of caution on proportion: divisibility is only a necessary condition; it can filter out the wrong ones but cannot sift out the right one: it can falsify, it cannot endorse.)

Three: the first four entries can be matched independently with the j-function. The coefficients of the j-function are defined by $j=E_4^3/\Delta$, have nothing whatsoever to do with the Monster, and can be computed completely independently: $196884$, $21493760$, $864299970$. Match them against the dimensions above:

$196884=1+196883,$ $21493760=1+196883+21296876,$ $864299970=2\cdot 1+2\cdot 196883+21296876+842609326.$

The three lines fit perfectly. This means the first four dimensions above are not “copied”: they have been independently checked once by a set of numbers from a completely unrelated source. As for why the coefficients of j are “exactly” sums of these dimensions: here we only lay out the arithmetic; that “why” is the business of the last episode. (This independent corroboration goes only up to the fourth entry; from the fifth entry on there is no such handle, and we can only cite the ATLAS.)

Four: this table was recomputed in 2024. First, one thing that is easily skipped over: the derivation of that original computation was never publicly published; Wilson himself recalled (in a December 2024 blog post) that “the calculations were so long and complicated, that nothing was ever published”. That is to say, for this table, cited over and over for more than forty years, the working draft was never in the public record.

As part of a project to “recompute all the character tables in the ATLAS”,

Breuer (Thomas Breuer), Magaard (Kay Magaard) and Wilson (Robert Wilson) redid the Monster’s table (arXiv:2412.12182). Their starting point is worth mentioning: under the assumption of the Monster’s uniqueness, plus the premise that “there exists a $196883$-dimensional faithful representation”, they determined all the conjugacy classes and the orders of the centralizers: the same machine as in $1979$ (exactly the column orthogonality proved above: the sum of squares of each column is exactly equal to the order of the centralizer of that class). Forty-five years on, walking the same set of constraints once more, it still goes through. And what they wrote at the start of the paper is:

“The table was checked in various ways before publication in the ATLAS in 1985, and its incorporation into GAP. Nevertheless, the problem of reliability and reproducibility of this table remains.”

Among the people who wrote this, Wilson is one of the five editors of the ATLAS, and Breuer is the author of the GAP character table library, that is, the library where this table is stored today. The people who compiled it back then and the people who keep it today, forty-five years later, looked back together to re-verify it, and said that the problem of reproducibility still exists. This is exactly another face of that EP5 motif: some things are too big to be verified completely by one pair of hands.

(Source of the dimensions: OEIS A001379 “Degrees of irreducible representations of Monster group M”: the data is generated by GAP’s CharacterTable("M") and originates in the ATLAS; the complete $194$ entries are in that entry’s b-file; $194$ entries / $170$ distinct, and the repetition in entries $16$–$17$, also come from that entry’s notes. The arithmetic of the three moonshine identities above, and “each entry divides $|M|$”, have been computed entry by entry for this episode.)

Why keep an eye on it (the hook to EP8). The epilogue of the main text says that the coefficient of the j-function is $196884=196883+1$: off by exactly a $1$. What is that $1$? Appendix C gives it a concrete identity right away.

Symbolic check: the $S_3$ table: first column = dimensions, $\sum n_i^2=1+1+4=6=|S_3|$, all pairs of rows weighted-orthogonal (including norm $=6$), all pairs of columns orthogonal and the sum of squares within a column equal column by column to the order of the centralizer of that class, $6/2/3$ (and checked independently against $|G|/|K|$), $r=k=3$; the counterexample to squareness (the $1\times2$ $A$ satisfies $AA^*=I_1$ while $A^*A\ne I_2$, checked cell by cell); the product of the prime factors of $|M|$ agrees digit by digit with the $54$-digit decimal (against the ATLAS); $47\cdot59\cdot71=196883$; the three largest in the set of prime factors of $|M|$ are exactly $\{47,59,71\}$; $196884=196883+1$. The proofs of Theorem 1 / Theorem 2 / $n_i\mid|G|$ are proofs in words ($S_3$ checked as their instance). That small piece of the real table: the $q$-expansion of the j-function is computed from scratch from $j=E_4^3/\Delta$ (an exact integer series), giving $196884$, $21493760$, $864299970$; the dimensions are cited from OEIS A001379 (its $\%o$ generated by GAP CharacterTable("M"), its $\%D$ pointing to ATLAS 1985); the three moonshine identities collide the computed j-coefficients with the cited dimensions, all equal ⟹ independent corroboration of the first four entries (only the first four entries; from the fifth on, cite-only). The first $18$ entries each divide $|M|$: necessary, not sufficient, not taken as an endorsement. The repeated entry $8{,}980{,}616{,}927{,}734{,}375=5^9\cdot7^6\cdot11^2\cdot17\cdot19$ occurs twice among the first $18$ entries (against the $\%C$ note’s $194$ entries / $170$ distinct). The Thompson order formula step: $|C_M(2\mathrm{B})|=2^{25}\cdot|\mathrm{Co}_0|/2=139{,}511{,}839{,}126{,}336{,}328{,}171{,}520{,}000$ ($|\mathrm{Co}_0|$ taken from EP6, already checked within the series); it divides $|M|$ (Lagrange, a real test); digit count $27$ vs $|M|$‘s $54$, exactly half, the same order of magnitude as $\sqrt{|M|}$. The formula itself, the definition of $a(x)$, the $194$ classes and the centralizer structure of 2A/2B, the sequence of dimensions, and $\mathbb{F}_2$‘s $196882$ = cited (Thompson’s original work / ATLAS / Breuer–Magaard–Wilson), not part of the check.