Composition Series and the Alternating Group

Much of finite group theory runs on a single strategy: find a normal subgroup $N$, understand $N$ and the quotient $G/N$, then reassemble. Both pieces are smaller than $G$, so the argument recurses. The obstruction is a group with no proper nontrivial normal subgroup, which cannot be broken down. Those are the simple groups, and every finite group decomposes into them uniquely.

Induction through a normal subgroup

The method is visible in a clean special case of Cauchy's theorem.

The proof inducts on $|G|$. Pick $x\ne 1$. If $p\mid |x|$, then a power of $x$ has order $p$. Otherwise set $N=⟨x⟩$ (normal, since $G$ is abelian); then $p\mid |G/N|$ and $|G/N|<|G|$, so induction gives an element of order $p$ in the quotient, which lifts back to one in $G$.¹ The shape recurs everywhere: information about $N$ and $G/N$ forces a conclusion about $G$, and the induction terminates because both are smaller.

The whole strategy stalls when $G$ has no usable normal subgroup.

Groups of prime order are simple, having no proper nontrivial subgroups at all. Every abelian simple group is $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$. There are non-abelian simple groups; the smallest has order $60$, appearing below as $A_5$.² Simple groups are the primes of group theory: unfactorable, and the building blocks of a unique factorization.

Composition series and Jordan-Hölder

Only $N_i⊴N_{i+1}$ is required, not $N_i⊴G$; the chain need not consist of subgroups normal in the whole group.

Every finite group has a composition series, and although the series itself is not unique, the multiset of factors is.

The proof is an induction using the second isomorphism theorem.³ The group $D_8$ shows the series is genuinely non-unique while the factors are not: $1⊴⟨s⟩⊴⟨s,r^2⟩⊴D_8$ and $1⊴⟨r^2⟩⊴⟨r⟩⊴D_8$ are different chains, but each has three factors, all isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

Counting all composition series is a lattice exercise. In $Q_8$ every maximal chain descends through one of the three order-$4$ subgroups $⟨i⟩,⟨j⟩,⟨k⟩$, then through the unique order-$2$ subgroup $\{\pm 1\}$: three composition series, all with factors ${\mathbb{Z}}_2,{\mathbb{Z}}_2,{\mathbb{Z}}_2$. In $D_8$ the middle level offers three subgroups of order $4$, and the two Klein-type ones each contain three subgroups of order $2$, giving $3+1+3=7$ composition series in total, again all with factor multiset $\{{\mathbb{Z}}_2,{\mathbb{Z}}_2,{\mathbb{Z}}_2\}$.⁴ Jordan-Hölder is visible in the count: many chains, one multiset.

The Hölder program and solvable groups

Jordan-Hölder splits the classification of finite groups into two problems, the Hölder program:

• Classify the finite simple groups. Completed around 1980 after roughly a century of work: every finite simple group is one of $18$ infinite families or one of $26$ sporadic groups.⁵
• Solve the extension problem. Given simple factors, describe all groups built from them. This is hard even for small factors, since nonisomorphic groups can share a composition-factor multiset ($Q_8$ and $D_8$ both have factors $\{{\mathbb{Z}}_2,{\mathbb{Z}}_2,{\mathbb{Z}}_2\}$).

One class named by its factors matters for Galois theory.

A finite group is solvable exactly when all of its composition factors have prime order.⁶ Solvability passes to subgroups and quotients, and is built by extension: if $N$ and $G/N$ are solvable, so is $G$. The extension argument applies the lattice isomorphism theorem directly. Take an abelian-quotient chain $1=N_0⊴\cdots ⊴N_n=N$ for $N$ and one for the quotient, $1={\overline{G}}_0⊴\cdots ⊴{\overline{G}}_m=G/N$. The lattice theorem lifts each ${\overline{G}}_i$ to a subgroup $G_i\le G$ containing $N$ with $G_i/N={\overline{G}}_i$, and the third isomorphism theorem converts each quotient of lifts back:

which is abelian. Concatenating the chain for $N$ with the lifted chain gives an abelian-quotient chain for all of $G$. The name solvable comes from the correspondence with polynomials solvable by radicals.

Transpositions and the sign homomorphism

The alternating group is defined by a parity invariant on permutations. A transposition is a $2$-cycle. Every permutation is a product of transpositions, via $(a_1{a}_2\cdots a_m)=(a_1{a}_m)(a_1{a}_{m-1})\cdots (a_1{a}_2)$ applied to each cycle, so $S_n=⟨T⟩$ for $T$ the set of transpositions. The factorization is far from unique, but its parity is fixed.

To make parity precise, act on the polynomial in $n$ variables

Each factor of $\sigma (\Delta )$ is $\pm (x_i-x_j)$, so $\sigma (\Delta )=\pm \Delta$. Define the sign $\epsilon (\sigma )=+1$ if $\sigma (\Delta )=\Delta$ and $\epsilon (\sigma )=-1$ if $\sigma (\Delta )=-\Delta$.

That $\epsilon (\tau \sigma )=\epsilon (\tau )\epsilon (\sigma )$ follows by tracking sign changes when $\tau$ permutes the factors of $\sigma (\Delta )$; computing $\epsilon$ on $(12)$ flips exactly one factor, and any transposition is conjugate to $(12)$, so all are odd.⁷

By the first isomorphism theorem $S_n/A_n\cong \{\pm 1\}$, so $A_n$ has index $2$ and $|A_n|=\frac{1}{2}n!$.

The sign is read directly from cycle structure. An $m$-cycle is a product of $m-1$ transpositions, so it is odd exactly when $m$ is even. Hence for any $\sigma$ with cycle decomposition ${\alpha }_1\cdots {\alpha }_k$, $\epsilon (\sigma )=\prod \epsilon ({\alpha }_i)$, giving a clean rule.⁸

The alternating group and simplicity

For small $n$ the alternating groups are familiar: $A_1$ and $A_2$ are trivial, $A_3=⟨(123)⟩\cong \mathbb{Z}/3\mathbb{Z}$, and $A_4$ has order $12$. The lattice of $A_4$ is a useful object: it has a unique subgroup of order $4$ (a copy of $V_4$, normal in $A_4$), four subgroups of order $3$, and three of order $2$, but no subgroup of order $6$, which is why $A_4$ is the standard counterexample to the converse of Lagrange.

The normal $V_4$ makes $A_4$ solvable, with composition series $1⊴⟨(12)(34)⟩⊴V_4⊴A_4$ and factors ${\mathbb{Z}}_2,{\mathbb{Z}}_2,{\mathbb{Z}}_3$, all of prime order. From $n=5$ on, the alternating groups are simple.

The proof, deferred to the study of group actions, turns on the $3$-cycles: they generate $A_n$, and a normal subgroup containing one $3$-cycle must contain all of them.⁹ Because $A_5$ is simple and non-abelian, $S_5$ is not solvable, and that single fact is the group-theoretic reason no formula in radicals solves the general quintic.