Symmetry and the Monster (3): The Indecomposable Atom · A₅

A word from the author

The proofs in this episode’s blog post were worked out step by step by me together with my AI partners. Whenever I had a question I kept pressing them, and of course they never let me down. Learning in the AI era may no longer need a teacher lecturing in a classroom, and may need more of our own desire to explore the unknown; and as knowledge becomes cheap and AI develops, one question will surely keep making us ask: what do we learn for? For me the answer is “I’m very curious: how on earth did this come about? Why?” For readers it may be different, but I hope readers will also find their own answers. Continuing from EP2 — in the last episode we set up group theory, asserted that “A₅ cannot be broken apart”, and left two IOUs (why it cannot be broken apart, and why being unbreakable means no formula can be written down), and also previewed a person who never met the other yet gave the same answer. This episode pays off all three at once. This is the episode in the series with the hardest mathematics, but it still

takes you as someone who has just finished the little group-theory lesson of the last episode, and goes down step by step from there. This episode’s promise:

Every conclusion that really matters in this episode is proved for you; the few standard lemmas found in every textbook are given with their sources and only their conclusions are taken, so as not to drown the main line.

“The chief forms of beauty are order, symmetry and definiteness, and the mathematical sciences display them in a special degree.” — Aristotle

(The chief forms of beauty are order and symmetry and definiteness, which the mathematical sciences demonstrate in a special degree. — Aristotle)

🎬 This episode’s companion video is now online (YouTube)

Contents

• An impossible game
• The letter that came too late
• §1 A three-hundred-year-old open case: Ruffini and Abel
• §2 The language of atoms: groups, composition series and Jordan–Hölder
• §3 Why A₅ is an atom
• §4 A little lesson in field theory: translating root extraction into groups
• §5 Closing the loop: the Abel–Ruffini theorem
• Epilogue: the first atom, and a list no one had seen

An impossible game

On the table there is a little toy: a four-by-four frame, fifteen sliding blocks, and one empty square left over. The blocks can slide horizontally and vertically; wherever the empty square moves, the block next to it can slide over. Scramble it, then try to slide it back into order — this is the “15 puzzle” that was all the rage in the nineteenth century.

Figure 1 The 14-15 puzzle illustration in Loyd’s 1914 Cyclopedia of 5000 Puzzles (public domain)

Now you are given a position that is almost set up: 1 to 15 arranged in order, except that the last two blocks — 14 and 15 have swapped places. The empty square is in the bottom-right corner. Your task: by sliding alone, swap 14 and 15 back, with the empty square back in the bottom-right corner again.

It sounds as simple as can be. And yet — you will find that you cannot do it no matter what.

This was exactly the prize problem of the day. America’s most celebrated puzzle master,

Sam Loyd (1841–1911)

, offered for it a

1000-dollar

reward (equivalent to well over a hundred thousand dollars today) for a solution.

He never had to worry about this reward — because the problem has no solution at all.

(A historical aside: this story of “Loyd’s invention + a thousand-dollar prize” is itself a widely circulated legend. Later research (Slocum & Sonneveld, 2006) established that the real inventor was Noyes Chapman, a postmaster in New York State, and that Loyd only began claiming publicly that it was his own creation in 1891. But

this takes nothing away from the mathematics of the game

— this impossibility is beyond doubt, and it is proved for you below.)

Why is it impossible? The answer is hidden in a plain but profound distinction: odd permutations and even permutations (in the little lesson of EP2 we saw “swaps” through the flips of a triangle; here we use them all the way).

First put the game aside, and think only about “scrambling”. The simplest scrambling is to swap two things and leave the rest alone — this is what EP2 called a swap (transposition). Six people sit around a table; two of them exchange seats and the others do not move: that is one transposition. Any complicated rearrangement can be carried out by breaking it into some number of transpositions.

Here there is a key fact (touched on at the end of the EP2 little lesson, and formally put to use here): if a rearrangement can be done with an even number of transpositions, it can never be done with an odd number, and vice versa. Every rearrangement is either “even” or “odd”, clearly one or the other. If you rearranged the seats with seven transpositions and someone says he can achieve the same result with exactly six, you can be sure: he is wrong.

🔑

Why are odd and even clearly separated?

(Put the people back one by one.) To get A to sit in B’s seat, swap A and B — A is back in place, and “the number of people in the right seat” goes up by at least one. In this way, each transposition flips the parity of the number of people sitting correctly, until everyone is back in place. The parity at the end point (all correct) is determined, so the parity of the number of transpositions needed to reach it is determined too.

Back to the game. Swapping 14 and 15 alone and leaving the rest alone — this is one transposition, an odd permutation. But any way of sliding that brings the empty square “back to the bottom-right corner” must be an even permutation. The reasoning is like coloring a chessboard: paint the 16 squares alternately black and white; each time the empty square slides one step, it goes from a black square to a white square, or from white to black — the color flips once. For the empty square to get back to its original square (the same color), the number of color flips must be even, that is, the total number of steps is even → an even permutation.

Odd ≠ even. An odd permutation can never be made up of an even number of steps. Loyd’s thousand dollars was therefore perfectly safe — and he knew this very well.

Figure 2 All arrangements of the 16 sliding blocks fall into two equal halves: the “even” half that can be reached by sliding and the “odd” half that cannot; “swap only 14 and 15” lands exactly in the unreachable half

By the way, a word on how “big” this game is: the full permutations of 16 positions give 16! = 20,922,789,888,000 arrangements (over twenty trillion); exactly half of them are even permutations, that is, 10,461,394,944,000 — this is the number of positions you can actually reach by sliding. The other half, more than ten trillion, including the one “swap only 14 and 15”, can never be reached by sliding.

This leap turns the 15-puzzle from a clever toy into the key to the whole episode: the “odd and even” you just played with is not an isolated special case. Shrink it to 5 objects — all the even permutations of 5 things number exactly $60$ and form a group, which is precisely the alternating group $A_5$ from the little lesson of EP2. And this $A_5$ cannot be broken apart; it is a “symmetry atom”; it is precisely the 60 rotations of the icosahedron that we ran into at the end of the last episode. That $A_5$ cannot be broken apart is also precisely the root of “the general quintic equation has no formula for its roots” — §5 will explain this line thoroughly.

In this episode, we fix our eyes only on this first and also smallest non-commutative atom, $A_5$: first prove that it really cannot be broken apart, then see how it leaves the quintic equation without a formula for its roots.

At the other end of the story are two young men who gave their lives to this atom.

The letter that came too late

Figure 3 Niels Henrik Abel (1802–1829), painted by Johan Gørbitz (public domain)

On April 6, 1829, in Norway, a 26-year-old young man’s life was taken by tuberculosis. Two days later — on April 8 — Crelle wrote from Berlin with good news: he had secured a position for him. The letter was two days late. His name was Niels Abel. Five years earlier (1824), he had printed at his own expense a thin little pamphlet proving what generations had failed to do: the general equation of the fifth degree has no formula for its roots (an expanded version was published in 1826 in the first issue of Crelle’s journal). But until his death he could not answer the next question — exactly which equations can be solved, and which cannot? That answer was in the hands of another person who was just as young, died just as early, and never met him — Galois. In this episode, we finish saying what the two of them did not have time to finish saying.

Figure 4 Évariste Galois (1811–1832), a portrait drawn from memory by his brother, at about age 15 (public domain)

§1 A three-hundred-year-old open case: Ruffini and Abel

The Babylonians could already handle quadratic equations; cubics and quartics were conquered in the Renaissance (covered in EP2); but the quintic equation held out against humanity for nearly three hundred years.

Figure 5 Paolo Ruffini (1765–1822), public domain

The first to declare solemnly that “it simply cannot be solved” was the Italian Paolo Ruffini: in 1799 he published his General Theory of Equations, asserting that the general equation of degree higher than four cannot have an algebraic solution, and he revised it repeatedly (1803, 1808, 1813). But his proof had a critical gap, and the mathematical world almost ignored it. He once asked the Paris Institute to rule on whether it was right or wrong — Lagrange, Legendre and Lacroix were assigned to review it, but the response was vague and disappointing. [According to the account in Ronan’s Symmetry and the Monster, the gist of the polite refusal at the time was: the geometers were unwilling to spend time delving into each other’s lengthy work.]

More than twenty years later, the Norwegian Abel (1824, independently of Ruffini) gave an accepted proof: there exist quintic equations whose solutions cannot be written in radicals. But note — x⁵=2 can be solved (just take the fifth root). So the real question is not “can the quintic equation be solved” but “which can, and which cannot”. Abel died of illness just as he was closing in on this measuring stick for deciding. The stage was left to Galois.

They had both touched the door, but could not push it open. To push it open, one must first think through, completely, the matter of “how groups break apart”.

§2 The language of atoms: groups, composition series and Jordan–Hölder

First list a “group” out in full: the Cayley table

Before proving anything, first make “the operation of a group” visible and checkable.

Take the symmetry group $S_3$ of the equilateral triangle from EP2. It has 6 actions: doing nothing $e$, rotating by $120°$, $r$, rotating by $240°$, $r^2$, and three flips $a,b,c$ (flipping about the axes through each of the three vertices). The operation of the group is “do one, then do another” — write $x\circ y$ = first do $y$, then do $x$ (the one on the right is done first, the EP2 convention).

Fill all the results of $x\circ y$ into a 6×6 table, and that is the multiplication table (Cayley table):

Figure 6 The Cayley table of S₃: each cell is the result of “row action ∘ column action”; the blue 3×3 block (gold frame) is exactly the subgroup A₃={e,r,r²}, closed in itself

This table turns the abstract “group” into a table you can check with your finger. Two things can be seen at a glance, and everything below depends on them:

• Each action appears exactly once in every row and every column (each row and each column is a rearrangement of the 6 actions) — this is a direct consequence of a group being “invertible”: fix $x$, and "$x\circ\,?$" sends different $y$ to different results, with nothing missed and nothing repeated.
• The blue 3×3 small block in the gold frame at the top left uses only $\{e,r,r^2\}$, and its results all fall in $\{e,r,r^2\}$ too — it is closed in itself, a little group of its own. This is a “subgroup”: a non-empty block of elements inside a group that is also a group on its own, with the door closed. Precisely, three conditions are needed: it contains the identity $e$, it is closed under composition $\circ$ (two things in the block multiplied are still in the block), and it is closed under taking inverses (the inverse of every element in the block is also in the block). (In a finite group, “non-empty + closed under $\circ$” in fact implies the other two; they are written out in full here to make later comparison easy.) This subgroup here is called $A_3=\{e,r,r^2\}$ (the three rotations).

Remember this table and this subgroup $A_3$; we are about to use it to prove the first theorem.

The first theorem: Lagrange (the size of a subgroup always divides the size of the group)

Lagrange (Joseph-Louis Lagrange, 1736–1813) was born in Turin, baptized Giuseppe Lodovico Lagrangia. His father had served as treasurer of the Office of Public Works and Fortifications in Turin, but lost the family fortune through failed speculation. Looking back in his later years, Lagrange actually counted this as a stroke of luck, and once said that if he had been rich at the start, he would probably not have devoted himself to mathematics. In the year he turned thirty, at the invitation of Frederick the Great of Prussia, he went to Berlin to succeed Euler as head of the mathematics section of the Academy of Sciences; it was precisely during his years in Berlin that he wrote Reflections on the Algebraic Solution of Equations (1770), the first systematic inquiry into “why equations of degree four and below can be solved by radicals, while those of degree five and above cannot” — this question, passing later through Ruffini and Abel, led straight to Galois, and became one of the sources of group theory.

In 1787 Lagrange moved to Paris. During the French Revolution and the Reign of Terror he escaped harm thanks to his high prestige and the protection of Lavoisier and others; when Lavoisier was sent to the guillotine, he lamented that it took only a moment to cut off that head, and a hundred years may not be enough to produce another like it. After Napoleon came to power, he was made a count of the Empire. But what truly keeps his name in group theory is a plain yet profound theorem. Take the most ordinary example — a square: its symmetries number $8$ in all (four rotations plus the flips about four axes of symmetry); among them the four rotations that “only turn and do not flip” form a subgroup of their own, and that $4$ divides $8$ exactly is no accident; this is a miniature of Lagrange’s theorem.

Theorem (Lagrange). For a finite group $G$, any subgroup $H$ of it has order dividing the order of $G$: $|H|$ divides $|G|$. (The notation $|\cdot|$ is read “how many elements are inside”, that is, the order of a group as EP2 put it — not an absolute value, it just borrows the same pair of vertical bars; e.g. $|S_3|=6$, $|A_3|=3$.) This is a proposition that holds for all finite groups, not a coincidence of some particular group.

Proof. For $g\in G$, write the left coset $gH=\{g\circ h:h\in H\}$ — the block obtained by left-multiplying the whole of $H$ by $g$, shifting it as a whole. The map $h\mapsto g\circ h$ is a bijection $H\to gH$ (with left multiplication by $g^{-1}$ as its inverse), so each coset has exactly $|H|$ elements. Moreover any two cosets are either equal or disjoint: if $x\in g_1H\cap g_2H$, write $x=g_1\circ h_1=g_2\circ h_2$; then $g_1=g_2\circ(h_2\circ h_1^{-1})\in g_2H$, so $g_1H\subseteq g_2H$, and symmetrically the reverse also holds, and the two blocks coincide. Therefore $G$ is tiled by a number of pairwise disjoint cosets each containing $|H|$ elements; let there be $m$ blocks in all, then $|G|=m\,|H|$, that is, $|H|$ divides $|G|$. $\blacksquare$

Check it with $S_3$: the subgroup $A_3=\{e,r,r^2\}$ and its coset $aA_3=\{a,b,c\}$, two blocks, fill it up, $3\times2=6=|S_3|$ ✓. So the sizes of the subgroups of $S_3$ can only be the divisors of $6$, $1,2,3,6$ — $4\nmid6$, so there cannot possibly be a $4$-element subgroup; Lagrange rules it out in one sentence.

This theorem is used everywhere later in this episode: A₅ has 60 elements, and the size of any subgroup of it (including “normal subgroups”) must divide 60. This is precisely one of the two locks that seal every gap when we prove “A₅ cannot be broken apart”.

We have just proved the first theorem — Lagrange’s theorem — from beginning to end. It is worth pausing here: theorems are the lifeline of mathematics. A theorem is a proposition that has been proved true; without them we would have nothing to rely on, because we would have no way of being sure whether the foundation under our feet is solid. If some conclusion looks true and we build it into a theory, but later find that it is false, then everything built on top of it collapses with it — every later result proved by means of it has to be torn down and redone. Precisely for this reason, mathematicians are meticulous about “proof”: an important conclusion that will be cited again and again must be proved until no one at all can find fault with it.

On this point mathematics is quite different from theoretical physics. The physicist Feynman has a widely circulated remark, whose gist is: the whole purpose of physics is to compute a number, with a decimal point — otherwise you have accomplished nothing. In mathematics, by contrast, the whole point is to state and prove a theorem.

Peeling a group layer by layer: composition series

In the EP2 little lesson we learned a trick: “shrink away a normal subgroup to get a quotient group”. First let us see this trick clearly with the Cayley table, then apply it repeatedly, all the way down.

First, a review of “quotient groups”, seen through the blocks of the Cayley table. Back to that table for $S_3$: $A_3=\{e,r,r^2\}$ (the blue block) is a normal subgroup, and the other block $\{a,b,c\}$ (the warm block) is its coset. Now treat each whole block as a new “point”: call the blue block “even” and the warm block “odd”. Look at the four corners of the Cayley table:

• blue block $\circ$ blue block = blue block (even∘even=even)
• blue block $\circ$ warm block = warm block (even∘odd=odd)
• warm block $\circ$ warm block = blue block (odd∘odd=even)

The whole 6×6 table shrinks into a 2×2 “multiplication table of blocks” — exactly the $\mathbb{Z}/2$ seen in EP2. This “multiplication left between points after each coset is pinched into a point” is the quotient group $S_3/A_3\cong\mathbb{Z}/2$. Quotient group = divide the big table into blocks by cosets, then look at how the blocks multiply.

A key precondition must be stated clearly: only when $N$ is a “normal” subgroup does this “block multiplication” make sense.

First recall what EP2 said about a

normal subgroup: take any $g$ and use it to move $N$, and moving it on the left, $gN$, and on the right, $Ng$, give the same block, that is, $gN=Ng$.

The whole use of normality is to make the definition “multiplying two blocks, $(gN)(g'N)=(gg')N$”, not depend on which element you pick from a block as its representative

— no matter whom you pick, the product lands in the same block. If $N$ were not normal, “block times block” would give different answers for different representatives, and would not be a definition at all. $A_3$ is normal in $S_3$ (checked in EP2), so $S_3/A_3$ makes sense.

Remember this: whenever a quotient group $G/N$ is written later, $N$ is assumed normal.

Composition series: peel until nothing more can be peeled. In EP2 we peeled one layer off $S_3$. Now peel layer by layer all the way down: find a normal subgroup and shrink it away, find another normal subgroup in what remains and shrink it away, until only $\{e\}$ is left. Each layer shrunk gives a quotient group. If the quotient at every layer can already ‘not be broken apart’ (it is a simple group as EP2 put it: a nontrivial group $G\ne\{e\}$ that, apart from $\{e\}$ and itself, has no other normal subgroup to shrink — by convention $\{e\}$ itself does not count as a simple group), this chain is called a composition series of $G$, and that string of simple-group quotients is the “symmetry atoms” of $G$.

An analogy: peel an onion layer by layer inward, until nothing more can be peeled; the layers peeled off are this onion’s “atomic layers”.

An example: peel $S_4$ all the way down. $S_4$ consists of the rearrangements of 4 objects $\{a,b,c,d\}$, all $4!=24$ of them. We peel it layer by layer, computing the quotient at each layer one by one:

First, two notations made clear: $G\triangleright N$ (triangle arrow) is read ”$N$ is a

normal subgroup

of $G$”; the arrow points from the big group to the block that is shrunk away; the string above is ”$S_4$ shrinks away $A_4$, $A_4$ then shrinks away $V_4$, shrinking all the way until only $\{e\}$ is left”. $\langle p\rangle$ (angle brackets) is read ”

the subgroup generated by $p$

” — the group made up of all the actions obtained by multiplying $p$ by itself repeatedly (EP2’s cyclic groups, generators).

Layer 1, $S_4/A_4$. $A_4$ is the 12 even rearrangements in $S_4$ (those built from an even number of swaps). As with $S_3$, even/odd each take half; $A_4$ is a normal subgroup, with one coset left over (all the odd rearrangements). Pinch by cosets into two points “even/odd”, and the quotient $=\mathbb{Z}/2$.

Layer 2, $A_4/V_4$. Here we need to lay $V_4$ out and see it clearly. It is these four actions in $A_4$:

(each is a double transposition, “swapping one pair in each of two groups”). It is itself a group — look at its Cayley table:

Figure 7 The Cayley table of V₄: the diagonal is all e (each flip multiplied by itself returns to the starting point), and two different flips multiplied give exactly the third → this is the Klein four-group Z/2×Z/2

One can see at a glance in the table: the diagonal is all $e$ (each action multiplied by itself = doing nothing), and any two different non-$e$ actions multiplied give exactly the third. This structure is called the Klein four-group, written $V_4\cong\mathbb{Z}/2\times\mathbb{Z}/2$.

First, this new notation "$\times$" made clear — the direct product (not covered in EP2).

The

direct product

$G\times H$ of two groups is just to put them

side by side, each minding its own business, not interfering with each other

. First look at the smallest, $\mathbb{Z}/2$: it is a “switch” — two states (off / on), press once to toggle, press twice to return to where it was. Then $\mathbb{Z}/2\times\mathbb{Z}/2$ is

two independent switches: a state is a pair $(\text{first},\text{second})$, $2\times2=4$ kinds in all — (off,off), (on,off), (off,on), (on,on); “multiplying” two states = the two switches each toggle independently, without affecting each other. Match these two switches to $V_4$: $e=$(off,off), $p=$(on,off), $q=$(off,on), $s=$(on,on). Check one: $p\circ q$ = “first on, second off” combined with “first off, second on” = the two switches together “first on, second on” = $s$ — exactly the $p\circ q=s$ read off the Cayley table above. Each action multiplied by itself returns to $e$ (press a switch twice and it resets), which also matches the diagonal being all $e$. So $V_4$ really and truly is “two independent copies of $\mathbb{Z}/2$”, genuinely $\cong\mathbb{Z}/2\times\mathbb{Z}/2$. (In general: the elements of $G\times H$ are pairs $(g,h)$, the operation is done component by component, $(g_1,h_1)(g_2,h_2)=(g_1g_2,\,h_1h_2)$, and the size is $|G\times H|=|G|\cdot|H|$. When larger direct products come up later, the meaning is the same: several groups side by side, each minding its own business.)

Why is $V_4$ normal in $A_4$? No new tool is needed, only “order + Lagrange”.

First, “the order of an element” made clear

(note: this is a different thing from the order of a group $|G|$ earlier): an action $x$ multiplied by itself repeatedly returns to $e$ sooner or later;

the smallest such number $k\ge1$ (with $x^k=e$) is called the order of $x$

. Example: in $A_4$ a double transposition multiplied by itself once $(x\circ x)$ returns to $e$, order $2$; a 3-cycle returns to $e$ after turning three times, order $3$; $e$ itself has order $1$.

The 12 actions of $A_4$ have orders only 1 ($e$), 2 (the three double transpositions), 3 (the eight 3-cycles); there is none of order 4. So any “subgroup of 4 elements” in $A_4$ cannot contain an element of order 4, nor can it contain an element of order 3 (if it did, it would have to have a 3-element subgroup, but $3\nmid4$ — $\nmid$ is read “does not divide” — violating Lagrange) — it can only be $\{e\}$ plus three elements of order 2, and the elements of order 2 are exactly those three double transpositions. So the only subgroup with 4 elements is $V_4$. A subgroup that is “one of a kind” must be normal: take any action $g$ and “relabel” $V_4$ as a whole (replace each $v$ by $gvg^{-1}$); what you get is still a 4-element subgroup in $A_4$ — but there is only one such subgroup, $V_4$, so it can only turn back into itself. This “unchanged under relabeling” is exactly normality. Hence $|A_4|/|V_4|=12/4=3$, and the quotient $=\mathbb{Z}/3$.

Layers 3 and 4. $V_4$ itself can be peeled further: it contains a normal subgroup $\langle p\rangle=\{e,p\}$ (of order 2; in a commutative group every subgroup is automatically normal). $V_4/\langle p\rangle$: $4/2=2=\mathbb{Z}/2$; finally $\langle p\rangle/\{e\}=\mathbb{Z}/2$.

⚠️

A place where it is easy to go wrong: you cannot stop at $V_4$! $V_4$ has

4

elements, not 2; it is

not

a simple group (it contains $\langle p\rangle$, which can be shrunk). So two more layers must be inserted. The composition series of $S_4$ therefore has

four layers, and the atoms are, in order,

$$\mathbb{Z}/2,\ \ \mathbb{Z}/3,\ \ \mathbb{Z}/2,\ \ \mathbb{Z}/2.$$

Check: $2\times3\times2\times2=24=|S_4|$ ✓. (Each time a layer is peeled, the size of the group is divided by the size of that layer’s quotient; peeled all the way down, the sizes of all the quotients multiplied together must restore the whole group — this is a quick way to check that nothing was peeled wrongly.)

At this point we know how to “peel a group into a string of atoms”. But at once there is a question that cannot be avoided: peeling another way, would we get a different set of atoms? We answer it below (the Jordan–Hölder theorem).

However you peel, the atoms are always the same set: Jordan–Hölder

Jordan (Camille Jordan, 1838–1922) worked in engineering early on — after graduating from the École Polytechnique in Paris he first made his living as an engineer, and only afterwards turned to mathematics, serving as examiner at the École Polytechnique and professor at the Collège de France. In 1870 he published Traité des substitutions et des équations algébriques (Treatise on Substitutions), the first monograph on group theory in history, which systematically organized Galois’s theory and remained the standard work on group theory for nearly thirty years afterwards, until Burnside’s book appeared. In the same year, Sophus Lie and Felix Klein made a special trip to Paris to study with him — each of the two later founded far-reaching directions in mathematics. In his later years Jordan suffered great grief: he had eight children (two daughters and six sons), and in the First World War three of his sons were killed one after another. In 1922 he died at the advanced age of 84.

Figure 8 Camille Jordan (1838–1922), public domain

What also keeps Jordan’s name in group theory is a theorem about “how to take a group apart”. A group can be peeled layer by layer inward like an onion; if this layer-by-layer peeling is likened to breaking matter down into molecules and further down into atoms, then what Jordan (and, a little later, the German mathematician Hölder) proved is: whichever path you follow to peel it, the set of “atoms” obtained when it is peeled all the way down is essentially always the same set. This is exactly the Jordan-Hölder theorem we are to prove in this section — it guarantees that the “composition factors” of a group do not depend on the way it is decomposed, which echoes precisely this series’ theme of “the atoms of symmetry”.

At the end of the “composition series” section we left a key question: peeling a group into a string of atoms, peeling another way, would we get a different set of atoms? This section and the next prove the answer, “no”. This is the hardest and most thought-demanding stretch of this episode; first we assemble two tools, then complete the proof.

The theorem to be proved (the Jordan–Hölder theorem): for any finite group $G$, however you peel its composition series, in whatever order, the set of composition factors (atoms) obtained at the end, together with how many times each kind of atom appears, is always the same set — differing at most in the order of arrangement. In other words, “which symmetry atoms a group breaks into” is a fingerprint inherent to and unique to that group, unrelated to how you take it apart. (This is a general theorem that holds for all finite groups, not a coincidence of some particular group.)

First, get a feel for what it says through an example (the example only helps understanding; it does not count as a proof): take $\mathbb{Z}/6=\{0,1,2,3,4,5\}$ (the 6 hours on a clock face, with the operation “add, then take the remainder mod 6”). It can be peeled all the way down in two different ways:

• Peeling A: $\mathbb{Z}/6\ \triangleright\ \{0,2,4\}\ \triangleright\ \{0\}$. First shrink away $\{0,2,4\}$ (this is a 3-element subgroup $\cong\mathbb{Z}/3$), and the remaining quotient shrinks the two piles “even/odd” each into one point $=\mathbb{Z}/2$; then shrink $\{0,2,4\}$ itself to $\{0\}$, quotient $=\mathbb{Z}/3$. Atoms: one $\mathbb{Z}/2$, one $\mathbb{Z}/3$.
• Peeling B: $\mathbb{Z}/6\ \triangleright\ \{0,3\}\ \triangleright\ \{0\}$. First shrink away $\{0,3\}$ (a 2-element subgroup $\cong\mathbb{Z}/2$), and the quotient shrinks the three piles “congruent mod 3” each into one point $=\mathbb{Z}/3$; then shrink $\{0,3\}$ to $\{0\}$, quotient $=\mathbb{Z}/2$. Atoms: one $\mathbb{Z}/3$, one $\mathbb{Z}/2$.

Two ways of peeling, and the atoms obtained are the same set $\{\mathbb{Z}/2,\ \mathbb{Z}/3\}$ (only in a different order). What J-H says is: this is by no means just a coincidence of $\mathbb{Z}/6$; it holds for all groups.

Figure 9 The two ways of taking Z/6 apart end up with the same set of symmetry atoms {Z/2, Z/3}

To prove it “for all groups”, two tools are needed — two isomorphism theorems. This section first explains the first one thoroughly (it is itself a generalization of that little EP2 example of “attaching plus and minus signs”), and later we prepare the second one and then close by induction.

Tool one: the first isomorphism theorem

First, using the “sign homomorphism $\operatorname{sgn}$” most familiar from EP2, get clear on three words — group homomorphism and its image and kernel — which will be used throughout what follows.

A group homomorphism is a “translator”. $\varphi:G\to H$ translates the “language” $G$ into the language $H$. It “preserves the operation” — the translation does not break the grammar: in $G$, “first do $a$, then do $b$”, translated over = in $H$, “first translate $a$, then translate $b$”, and the result is the same.

Use as the example the translator from EP2 you know best: attaching plus and minus signs, $\operatorname{sgn}$. It translates $S_3$ (the 6 symmetry actions of the equilateral triangle, $e,r,r^2,a,b,c$) into the group with only two numbers, $\{+1,-1\}$ (operation = multiplying the two numbers): the three rotations are translated to $+1$, and the three flips to $-1$. Listed one by one:

It does indeed “preserve the operation” (for example “flip ∘ flip = rotation” corresponds to "$(-1)\times(-1)=+1$"). Keeping our eyes on this translator, let us learn two words.

Image = the places the translation actually reaches. Translate all the actions of $S_3$ over; the set made up of the values actually reached in the target is called the image, written $\operatorname{Im}\varphi$. $\operatorname{sgn}$ actually only puts out $+1$ and $-1$, so

(Here it happens to cover the whole target group $\{+1,-1\}$, which is called surjective; if some value were never produced by the translation, the image would not be the whole target.)

Kernel = those originals that are translated into “the identity” (the translation says nothing at all). The identity of the target group $\{+1,-1\}$ is $+1$ (multiplying anything by it leaves it unchanged = “saying nothing”). Which actions are translated to $+1$? Look at the table above — the three rotations. So

Kernel = “how much information” this translator “loses”. $\operatorname{sgn}$ squeezes 6 actions into 2 signs; the 3 actions in the kernel $A_3$ are all translated into the same $+1$, and can no longer be told apart from one another — this is “losing information”, and the bigger the kernel, the more is lost. Conversely, a translator whose kernel is only $\{e\}$ (only “doing nothing” is translated into the identity) loses nothing at all, and different originals are translated into different translations = injective (example: the “identity translator” that translates each action into itself has kernel $\{e\}$). The kernel of $\operatorname{sgn}$, $A_3$, has 3 elements; it is a translator that “loses quite a lot”.

Now connect it to that big theorem. Keeping our eyes on $\operatorname{sgn}$: kernel $=A_3$ means “the three rotations in $A_3$ all come out the same when translated (all are $+1$)”. We computed earlier: $S_3$, cut into two blocks by $A_3$ (the three rotations one block, the three flips another), pinched into two points, is the quotient group $S_3/A_3$, whose multiplication is exactly $\mathbb{Z}/2$. And the image $\{+1,-1\}$, as a group, is also $\mathbb{Z}/2$. Two blocks ↔ two translations, in one-to-one correspondence, with the same skeleton:

Here it is $\cong$, not an equals sign $=$

(it is the earlier $\ker\operatorname{sgn}=A_3$ and $\operatorname{Im}\operatorname{sgn}=\{+1,-1\}$ that use a true equals sign). $=$ governs “literally the same pile of things”; $\cong$ (read ”

isomorphic

”, covered in EP2) governs ”

as groups, the skeleton of multiplication is exactly the same, only the names differ

”. Here the left side is two

points, “rotation block / flip block”, and the right side is two

numbers, ”$+1$ / $-1$”; literally they are not the same thing, only the skeleton of multiplication is the same — so $\cong$ is used. Keep $=$ (literally the same) and $\cong$ (the same skeleton) apart, and there is no ambiguity.

This phenomenon is not unique to $\operatorname{sgn}$; it holds for any translator:

So, given any group homomorphism (translator) $\varphi:G\to H$, we have

In plain words: merge the original group by its “kernel” (the pile translated into nothing), and what is left has exactly the same skeleton as “all the translations” — the translation drops everything it should drop, and what remains is a clean one-to-one correspondence.

Why? The key is a single sentence: the originals that are translated into the same translation form exactly one coset of the kernel — what “merging” merges is precisely that.

On $\operatorname{sgn}$ it can be seen very clearly: the whole of what is translated to $+1$ $=\{e,r,r^2\}=A_3=$ the kernel (the coset containing the identity); the whole of what is translated to $-1$ $=\{a,b,c\}=$ another coset of the kernel. Two translations ↔ two cosets, in one-to-one correspondence. And the cosets pinched into points are exactly the quotient group $G/\ker$, while the whole of the translations is exactly the image — the two naturally have the same skeleton. Writing this sentence rigorously is the proof:

Intuition: the elements mapped to the same value form exactly one coset of the kernel, so the quotient group obtained by merging $G$ by the kernel, $G/\ker\varphi$, corresponds one-to-one, and preserving the operation, with all the values taken, $\operatorname{Im}\varphi$. $\operatorname{sgn}$ is a special case of it: $S_3/A_3\cong\{+1,-1\}\cong\mathbb{Z}/2$. The way this theorem is used: when you meet a quotient group $G/N$ that is hard to identify, just build a homomorphism whose kernel is exactly $N$, and $G/N$ is isomorphic to its image — the second isomorphism theorem below uses it in exactly this way.

Finishing the proof of Jordan–Hölder by diamond induction

Earlier we prepared tool one (the first isomorphism theorem). This section prepares tool two and one small fact, and then proves J-H completely.

Tool two: the second isomorphism theorem

Statement: Let $A$, $B$ both be subgroups of $G$, and let $B$ be normal in $G$. Then $A\cap B$ is normal in $A$, and moreover

⚠️

$AB$ is not a direct product; let us be clear about that first.

$AB$ means the set made up of all the results obtained by “taking an element of $A$ and an element of $B$ and

multiplying them in the group $G$

”, $\{a\circ b:a\in A,\,b\in B\}$ — its elements are all, inside $G$,

ordinary elements, and as a whole it is a subset of $G$ (and, when $B$ is normal, a subgroup too). Whereas the

direct product

$A\times B$ (the kind in $V_4=\mathbb{Z}/2\times\mathbb{Z}/2$ earlier) is a completely different thing: its elements are

pairs

$(a,b)$, and it is a newly constructed group. The two also differ in size — $|AB|=\dfrac{|A|\,|B|}{|A\cap B|}$, while $|A\times B|=|A|\,|B|$. What the second isomorphism theorem uses here is the

product set $AB$

(living inside $G$), not the direct product.

Proof: apply the first isomorphism theorem to the map $\psi:A\to AB/B,\ a\mapsto aB$, and it follows — $\psi$ is a surjective homomorphism, and its kernel is exactly $A\cap B$.

A small fact: the subgroup at each step of a composition series is “maximal normal”

A composition series requires each quotient to be a simple group. There is an equivalent way of saying this, which the induction will need:

Small fact: If the quotient group $G/N$ is a simple group, then $N$ is a maximal normal subgroup of $G$ — meaning that between $N$ and $G$ no other normal subgroup can be squeezed in (apart from $N$, $G$ themselves).

Proof

(only the direction needed): if there is a normal subgroup $M$ squeezed in between, $N\subsetneq M\subsetneq G$, then $M/N$ is a nontrivial proper normal subgroup of $G/N$, contradicting $G/N$ being a simple group. So $N$ must be maximal normal.

The induction: finishing the proof of Jordan–Hölder

The tools are all here. We want to prove: for any finite group $G$, any two composition series of it give the same set of atoms (composition factors) (with multiplicities, differing at most in arrangement).

Use “induction on $|G|$” — that is, assume “it already holds for every group smaller than $G$”, and prove that it holds for $G$ too. Base case: when $|G|=1$ (the group has only $e$), there is only the one trivial composition series, nothing to compare, and it holds automatically. Below, let $|G|>1$, and assume it already holds for groups smaller than it. Let $G$ have two composition series

Their first atoms are $G/G_1$ and $G/H_1$ respectively. There are two cases.

Case one: the two chains peel off the same subgroup at the first step ($G_1=H_1$). Then from the second step on, both chains are composition series of the same smaller group $G_1$. $G_1$ is smaller than $G$, so by the induction hypothesis the atoms of these two chains of it are the same set. Adding the common atom at the very top, $G/G_1$, the whole thing matches. ✓

Case two: what is peeled off at the first step is not the same ($G_1\ne H_1$). This is the key step, and it relies on that diamond diagram:

Figure 10 The diamond of the Jordan–Hölder induction: G at the top, the two maximal normal subgroups G₁/H₁ on the two sides, their intersection K at the bottom; the second isomorphism theorem gives that the quotients on opposite sides are isomorphic to each other, G/G₁≅H₁/K, G/H₁≅G₁/K

• Since $G/G_1$, $G/H_1$ are both simple groups, by the small fact above, $G_1$ and $H_1$ are both maximal normal subgroups of $G$ (the two below the top vertex $G$ in the figure).
• Look at their product $G_1H_1$. First, it is a normal subgroup (the product of two normal subgroups is still normal: $g(G_1H_1)g^{-1}=(gG_1g^{-1})(gH_1g^{-1})=G_1H_1$). Next, it properly contains $G_1$ — this step needs care: knowing only $H_1\ne G_1$ is not enough (in case $H_1$ were a proper subset of $G_1$, it would not bring a single new element, and $G_1H_1$ would still be $G_1$). The real reason is the maximality of $H_1$: suppose $H_1\subsetneq G_1$; then $H_1\subsetneq G_1\subsetneq G$ squeezes a normal subgroup $G_1$ between $H_1$ and $G$, contradicting ”$H_1$ is maximal normal (nothing else can be squeezed in between)”. So $H_1\not\subseteq G_1$ — there really is an element in $H_1$ that is not in $G_1$, and it adds to $G_1H_1$ cosets outside $G_1$, so $G_1H_1\supsetneq G_1$. Then, using the maximality of $G_1$ (between $G_1$ and $G$ no other normal subgroup can be squeezed in either), the $G_1H_1$ squeezed in between can only be $G$ itself. (Top of the figure: $G_1H_1=G$.)
• Let $K=G_1\cap H_1$ (bottom of the figure: the intersection of the two). Apply tool two (the second isomorphism theorem) to the two diagonals of this diamond: $$\underbrace{G/G_1}_{\text{upper-left edge}}=G_1H_1/G_1\cong \underbrace{H_1/K}_{\text{lower-right edge}},\qquad \underbrace{G/H_1}_{\text{upper-right edge}}\cong \underbrace{G_1/K}_{\text{lower-left edge}}.$$ The two quotients on opposite sides of the diamond are isomorphic to each other. This is the core of the whole proof.
• Since $G/G_1$, $G/H_1$ are simple, by the isomorphisms just obtained $H_1/K$, $G_1/K$ are simple too — and then by the small fact, $K$ is maximal normal both in $G_1$ and in $H_1$. For $K$, take any composition series $K\triangleright\cdots\triangleright\{e\}$.

So, attaching any composition series to $K$, we get two new chains passing through $K$, $G\triangleright G_1\triangleright K\triangleright\cdots$ and $G\triangleright H_1\triangleright K\triangleright\cdots$: by the diagonal isomorphisms, the pairs of atoms at the top of the two are the same up to renaming, and below them follows the same $K$ chain, so their atoms are the same set. Then, using the induction hypothesis on each, align original chain A with “the new chain via $G_1$”, and original chain B with “the new chain via $H_1$”, equal all the way:

The atoms of the two original composition series are the same set. The Jordan–Hölder theorem is proved. $\blacksquare$

Looking back, a one-sentence summary of the cleverness here: the two ways of peeling diverge at the first step ($G_1\ne H_1$), but their

intersection $K$

joins the two paths together again; the second isomorphism theorem guarantees that when “coming down around the left” and “coming down around the right” meet at $K$,

the atoms picked up along the way are the same set, only with their order swapped

. That example of $\mathbb{Z}/6$ ($\{0,2,4\}$ and $\{0,3\}$ diverging, meeting at $\{0\}$, the atoms $\{\mathbb{Z}/2,\mathbb{Z}/3\}$ swapped) is the smallest physical instance of this diamond.

Full proofs

— for the first and second isomorphism theorems and the induction of the Jordan–Hölder theorem, see Dummit–Foote, Abstract Algebra, 3rd ed. (Wiley 2004) §3.3–§3.4; the historical source for composition series and the Jordan–Hölder theorem is Jordan 1870, Traité des substitutions.

With this, “which symmetry atoms a group breaks into” is its unique fingerprint — proved. With this measuring stick in hand, the next section (§3) returns to the atom that stands in the way: what makes $A_5$ impossible to break apart?

§3 Why A₅ is an atom

What makes A₅ impossible to break apart (part 1 · conjugation → conjugacy classes → normal = union of whole classes)

With J-H’s measuring stick of “atoms are unique” in hand, we return to the atom that stands in the way, $A_5$. In EP2 we asserted that it cannot be broken apart; now we must really prove it.

First, the lettering convention for this stretch made clear: from here on, $A_5$ is

the even permutations of 5 objects, and we call these 5 objects $a,b,c,d,e$ (not $1,2,3,4,5$; numbers are easily confused with “counts/orders”). Permutations are written as cycles: $(ab)$ means “swap $a,b$ and leave the rest alone”, $(abc)$ means "$a\to b\to c\to a$", $(abcde)$ means the 5 arranged in a ring and rotated. (Note: in the $S_3$ of the EP2 little lesson, $a,b,c$ were the names of “three flip actions”; here $a,b,c,d,e$ are “the 5 objects being scrambled” — different contexts, do not mix them up.)

The theorem to be proved: $A_5$ ($|A_5|=60$) has no nontrivial normal subgroup — apart from $\{e\}$ and itself, no third normal subgroup that can be “shrunk away as a whole block” can be found. (So it cannot be broken apart = a simple group.)

To prove it we need a new tool that EP2 did not formally cover — conjugation and conjugacy classes. Below we start from a familiar small group.

Conjugation: changing to a different numbering, looking at the same action

EP2 said that a subgroup $N$ being normal means “moving it left and right gives the same thing”: take any element of the group $g$ ($g$ is just any action/permutation in this group; whichever one you pick is fine), and we always have $gN=Ng$. Right-multiplying both sides by $g^{-1}$ gives an equivalent way of writing it:

The operation that pops up here, "$x\mapsto gxg^{-1}$" (take the $g$ you picked and sandwich $x$ between it front and back), is called using $g$ to conjugate $x$. What is the intuition? $g$ amounts to giving all the objects a new set of numbers, and $gxg^{-1}$ is “under the new numbering, looking again at the same action $x$”.

It is clearest on permutations — the role of $g$ is exactly that “renumbering”.

Take a permutation as an example: $\sigma=(a\,b)$, meaning “swap $a,b$ and leave the rest alone”. Now pick a $g$ to serve as the rule for renumbering — for example $g=(a\,b\,c)$, whose effect is to

rename $a$ as $b$, rename $b$ as $c$, rename $c$ as $a$

(this is $g(a)=b,\ g(b)=c,\ g(c)=a$). Then “using this $g$ to conjugate $\sigma$” is

renaming each letter in the cycle of $\sigma$ once according to $g$:

$$g\,\underbrace{(a\ \ b)}_{\sigma=\text{swap }a,b}\,g^{-1}\ =\ \big(\ \underbrace{g(a)}_{=\,b}\quad \underbrace{g(b)}_{=\,c}\ \big)\ =\ (b\ \ c).$$

In plain words: originally it was “swap $a,b$”; after the two names $a,b$ are changed according to $g$ into $b,c$, it becomes “swap $b,c$”.

The same kind of action (both are a single swap), only with a different set of names for the objects.

(If you are not reassured, you can check $g\sigma g^{-1}$ point by point: acting on $b$ — first $g^{-1}$ turns $b$ back into $a$, then $\sigma$ swaps $a$ into $b$, then $g$ renames $b$ as $c$; net effect $b\mapsto c$; likewise $c\mapsto b$, $a\mapsto a$, which together are exactly $(b\ c)$.)

The general rule

(remember this one): $g\,(x_1\,x_2\,\cdots\,x_k)\,g^{-1}=\big(g(x_1)\ g(x_2)\ \cdots\ g(x_k)\big)$ — conjugation is “renaming the letters in the cycle uniformly by $g$”,

it does not change the shape of the cycle, it only changes the frame of reference

.

Conjugacy classes: a family of actions that “coincide with each other once the numbering is changed”

Conjugate an action $\sigma$ by each of all the $g$ in the group (with the renaming rule of the previous section); all the results together make up a conjugacy class — a family of actions that “can coincide with each other as long as the objects are given a different numbering”.

First, bring conjugacy classes down to a concrete example on $S_3$. Write $S_3$ as all 6 permutations of 3 objects $a,b,c$: the identity $e$, three swaps $(ab),(ac),(bc)$, two 3-cycles $(abc),(acb)$. Conjugate each one by all $g$, and you will find that they automatically fall into three classes according to “whether the shape is the same”:

• $\{e\}$: however $e$ is conjugated it is still $e$; it is a class of its own.
• $\{(ab),(ac),(bc)\}$: the three swaps are conjugate to each other — the previous section computed $(abc)\,(ab)\,(abc)^{-1}=(bc)$; changing to another $g=(bc)$ can also turn $(ab)$ into $(ac)$ ($(bc)\,(ab)\,(bc)^{-1}=(ac)$). The three swaps are one class.
• $\{(abc),(acb)\}$: the two 3-cycles are conjugate to each other, one class.

$6=1+3+2$: $S_3$ splits into 3 conjugacy classes, and exactly according to “shape”. This “shape” can be stated precisely, and so there is a criterion. First let us state the definitions of two words clearly.

Definition (cycle type). Fix $n$ objects. Any permutation can be broken uniquely into several mutually non-overlapping cycles (EP2 used this; an object that does not move counts as a cycle of length 1). The cycle type of a permutation is the shape formed by the lengths and numbers of its cycles after it is broken into disjoint cycles — looking only at “how many cycles, and how long each is”, not at which particular letters (counting the objects that do not move as cycles of length 1, the lengths of all the cycles always add up to the total number of objects $n$). Examples (on 5 objects): the shape of $(ab)$ is “one swap + three fixed points”; $(ab)(cd)$ is “two swaps + one fixed point”; $(abc)$ is “one 3-cycle + two fixed points”. — The “swap / 3-cycle / fixed point” here all describe the lengths and numbers of cycles, not which particular objects (the objects are always the letters $a,b,c,d,e$).

Definition (conjugate, as a relation between two permutations). To say that $\sigma$ and $\tau$ are conjugate in a group $G$ means: there exists in the group a renaming $g\in G$ such that $g\sigma g^{-1}=\tau$. In plain words, “changing to a different numbering can turn $\sigma$ into $\tau$”. ($g\sigma g^{-1}$ is the conjugation operation defined in the previous section; now we give the fact ”$\sigma$ can be turned into $\tau$ in this way” the name “conjugate”.)

First, set out the “renaming rule” we will need as a named fact (the previous section gave it; here we make clear why it is true, so that this section is self-contained):

Renaming rule: $g\,(x_1\,x_2\,\cdots\,x_k)\,g^{-1}=\big(g(x_1)\ g(x_2)\ \cdots\ g(x_k)\big)$; with several cycles, rename each one this way.

Why: under the convention that the right one is done first, $g\sigma g^{-1}$ acting on $g(x)$ = first $g^{-1}$ goes back to $x$, then $\sigma$ sends it to $\sigma(x)$, then $g$ sends it to $g(\sigma(x))$; that is, it sends $g(x)$ to $g(\sigma(x))$. $\sigma$ originally sends $x$ to $\sigma(x)$, so $g\sigma g^{-1}$ is nothing more than “renaming every letter in $\sigma$ according to $g$” — the shape has not moved at all.

Criterion (in the full symmetric group $S_n$): $\sigma$ and $\tau$ are conjugate $\iff$ they have the same cycle type. The two directions are proved separately.

($\Rightarrow$ conjugate ⟹ same type; this half holds in any group.) Let $\tau=g\sigma g^{-1}$. By the renaming rule, $\tau$ just renames the letters in each cycle of $\sigma$ according to $g$: a cycle of length $k$ after renaming is still of length $k$, and the number of cycles, their respective lengths, even the number of fixed points — not one of them has changed. So $\tau$ and $\sigma$ have the same cycle type.

($\Leftarrow$ same type ⟹ conjugate; this half needs “the renaming $g$ to lie in the group”.) Let $\sigma,\tau$ have the same cycle type. Pair up the cycles on the two sides by length, and align them position by position: match the first letter in some cycle of $\sigma$ to the first letter in the corresponding cycle of $\tau$, the second to the second… the letters that $\sigma$ fixes are also matched to the letters that $\tau$ fixes. Because the two sides have the same cycle type (fixed points included, the lengths add up to $n$ on both), this alignment matches $\sigma$‘s $n$ letters with $\tau$‘s $n$ letters one to one, with nothing repeated and nothing missed — so this “letter to letter” correspondence is a bijection on $n$ objects, that is, a genuine permutation $g$. By the renaming rule, it gives exactly $g\sigma g^{-1}=\tau$. In $S_n$, $g$ is just some permutation of these $n$ objects, and necessarily belongs to $S_n$, so $\sigma,\tau$ are conjugate in $S_n$. (Note: such a $g$ exists, but is in general not unique — changing the starting point at which each cycle is aligned, or changing how cycles of equal length are paired, gives another legitimate $g$. This point will be needed later for $A_5$.) Proved. $\blacksquare$

Worked examples (go through constructing the $g$ for ⟸ twice). 1. $\sigma=(ab)$, $\tau=(bc)$ (the same type, both one swap + one fixed point). Align: $a\to b$, $b\to c$; the remaining letter that $\sigma$ fixes, $c$, can only be matched with the letter that $\tau$ fixes, $a$, that is, $c\to a$. Together $g=(abc)$. Substitute: $g\,(ab)\,g^{-1}=\big(g(a)\ g(b)\big)=(b\,c)=\tau$ ✓. 2. A longer one, to see “aligning starting points + fixed points” clearly: $\sigma=(abc)$, $\tau=(bcd)$ (on 4 objects $a,b,c,d$, the same type, both one 3-cycle + one fixed point). Align $a\to b,\ b\to c,\ c\to d$; the letter $\sigma$ fixes, $d$, is matched with the letter $\tau$ fixes, $a$, that is, $d\to a$. Together $g=(abcd)$. Substitute: $g\,(abc)\,g^{-1}=\big(g(a)\ g(b)\ g(c)\big)=(b\,c\,d)=\tau$ ✓.

So in $S_n$, conjugacy classes = exactly the classification by cycle type; that is where those three classes of $S_3$, $1+3+2$, come from.

⚠️

Take special note: once inside a subgroup (such as the $A_5$ we want to prove things about), this criterion changes.

The step “same type ⟹ conjugate” requires “the renaming $g$ to be in the group”. But in $A_5$, those $g$ that can rename two permutations of the same type onto each other

might all be odd permutations, with not a single one in $A_5$

— in that case the same cycle type in $A_5$

splits into more than one

conjugacy class. (More precisely: it really splits only when

no even permutation at all

can rename the two onto each other; this precise judgment — which in fact amounts to looking at whether the “centralizer” of $\sigma$ consists entirely of even permutations — is left until later in this section, where it is worked out with centralizers.)

The 24 5-cycles of $A_5$ split in exactly this way, into two classes of 12 each

. So the conjugacy classes of $A_5$

are not

the 4 classes $\{1,15,20,24\}$ given by cycle type, but

5 classes

$\{1,15,20,12,12\}$. This is, in the whole $A_5$ proof,

the most critical point: overlook one class, and the case-by-case check later will miss a case, and the proof will not be rigorous.

The key step: a normal subgroup = several whole classes put together

Now connect conjugation and normality. Look back at the definition of normal, $gNg^{-1}=N$ — what it says is “after each element of $N$ — call it $x$ — is renamed (conjugated) into $gxg^{-1}$, the whole of $N$ is left untouched”. This implies a very strong conclusion:

As soon as a normal subgroup $N$ contains some action $x$, it is forced to contain all the conjugates of $x$ too (that is, the whole conjugacy class that $x$ is in).

Why: normality requires $gxg^{-1}$ (for

every

$g$) to still land in $N$; and what $x$ sweeps out when conjugated by all $g$

is exactly the conjugacy class of $x$

. So containing $x$ means containing its whole class.

See what this sentence means on $S_3$. Suppose a normal subgroup $N$ contains the swap $(ab)$. By the statement above, $N$ is forced to take in the whole conjugacy class of $(ab)$, $\{(ab),(ac),(bc)\}$ (not one of the three swaps may be missing). Adding the $e$ it must contain, $N$ has at least these 4: $\{e,(ab),(ac),(bc)\}$. But $4\nmid 6$ (Lagrange: the order of a subgroup must divide the order of the group, $6$)! So such an $N$ cannot stop at 4, and can only expand all the way to fill the whole of $S_3$. Conclusion: in $S_3$, as soon as a normal subgroup contains one swap, it has to be the whole of $S_3$.

Conversely, $A_3=\{e,(abc),(acb)\}$ is a proper, legitimate normal subgroup — it is exactly “the class of $\{e\}$ + the class of 3-cycles”, two whole classes put together ($1+2=3$, and $3\mid6$ ✓, not violating Lagrange). The “recipe” for a normal subgroup is: put it together from whole conjugacy classes, and the total must also divide the order of the group. In general:

This turns the hard job of “finding normal subgroups” into “choosing several complete conjugacy classes to put together, while also satisfying Lagrange (the total divides the order of the group)” — and how big each class is, and how many classes there are in all, can be counted. Later in this section we count the 5 conjugacy classes of $A_5$, list all 60 elements, and then put them together and check in this way, concluding that ”$A_5$, apart from $\{e\}$ and itself, has no other normal subgroup = cannot be broken apart”.

What makes A₅ impossible to break apart: lay out the 60 elements and check them class by class

Earlier we got a measuring stick: any normal subgroup is a union of several complete conjugacy classes (and always contains the class of $e$); and by Lagrange, its size must divide the order of the group. Now we measure $A_5$ with this stick — first count its conjugacy classes and lay out all 60 elements, then check them one by one.

1. The 60 elements of $A_5$, all listed by the 5 conjugacy classes

$A_5$ is all the even permutations of the 5 objects $a,b,c,d,e$ (those built from an even number of swaps), 60 in all. By conjugacy class they fall into 5 classes (the first 3 classes are just by shape; the 5-cycles need care, see “2. Centralizers” below):

Figure 11 The 5 conjugacy classes of A₅ counted directly: identity 1, double transpositions 15, 3-cycles 20, 5-cycles split into 12+12, total 1+15+20+12+12=60

Class 1 · the identity $e$ (1 element). Nothing moves.

Class 2 · double transpositions of type $(ab)(cd)$ (15 elements). Two groups, each swapping one pair, with one object left fixed. How to count: first choose which object is fixed (5 ways); the remaining 4 can be divided into two pairs in 3 ways, $5\times3=15$. (Each is two swaps = an even permutation.) The full list:

Class 3 · 3-cycles of type $(abc)$ (20 elements). Three objects rotate, two are left fixed. How to count: choosing 3 objects has $\binom{5}{3}=10$ ways, and each group arranged in a ring has 2 directions ($(abc)$ and $(acb)$), $10\times2=20$. (One 3-cycle = two swaps = an even permutation.) The full list:

Classes 4 and 5 · 5-cycles of type $(abcde)$ (24 in all, but split into two classes of 12 each). Five objects arranged in a ring and rotated, with no fixed point. How to count: arranging 5 objects in a ring has $5!$ arrangements, but the same ring written from its 5 starting points is the same, $5!/5=24$. (One 5-cycle = four swaps = an even permutation.) These 24 are not one class, but split into two classes of 12 each — why they split, and how they divide, is exactly the ⚠️ left earlier, worked out precisely below with “centralizers”. The two classes of 12 each, listed in full (the results are laid out first; the proof of the split is below in “2. Why the 5-cycles split into 12+12”):