---
title: Composition Series and the Alternating Group
module: Subgroups and Quotients
moduleNumber: 3
lessonNumber: 6
order: 306
summary: >
  A composition series breaks a finite group into simple quotient factors, and
  Jordan-Hölder says those factors are unique up to order. This turns
  classification into two problems: list the simple groups, and describe how to
  reassemble them. The sign homomorphism splits $S_n$ into even and odd
  permutations, defining the alternating group $A_n$, simple for $n \ge 5$.
topics: [Subgroups and Quotients]
sources:
  - book: Dummit & Foote
    ref: "Ch. 3; §3.4 Composition Series and the Hölder Program, §3.5 Transpositions and the Alternating Group"
  - book: Judson
    ref: "Ch. 10 Normal Subgroups and Factor Groups; §10.2 The Simplicity of the Alternating Group"
draft: false
---

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.

> **Proposition.** If $G$ is a finite abelian group and a prime $p$ divides
> $|G|$, then $G$ has an element of order $p$.

The proof inducts on $|G|$. Pick $x \neq 1$. If $p \mid |x|$, then a power of $x$
has order $p$. Otherwise set $N = \langle x \rangle$ (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 $G$.[^df-cauchy] 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.

> **Definition (Simple group).** A group $G$ is **simple** if $|G| > 1$ and its
> only normal subgroups are $1$ and $G$.

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
$A_5$.[^df-simple] Simple groups are the primes of group theory: unfactorable,
and the building blocks of a unique factorization.

## Composition series and Jordan-Hölder

> **Definition (Composition series).** A **composition series** for $G$ is a
> chain
> $$
> 1 = N_0 \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_k = G
> $$
> in which each quotient $N_{i+1}/N_i$ is simple. The quotients $N_{i+1}/N_i$
> are the **composition factors** of $G$.

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

$$
% caption: A composition series climbs from $1$ to $G$ through subgroups, each
% normal in the next, with every successive quotient a simple group.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  nd/.style={draw, inner sep=3pt, fill=acc!8, minimum width=13mm, font=\small}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[nd] (n0) at (0,0)   {$1$};
  \node[nd] (n1) at (0,1.2) {$N_1$};
  \node[nd] (n2) at (0,2.4) {$N_2$};
  \node[black] (dots) at (0,3.3) {$\vdots$};
  \node[nd, draw=acc] (nk) at (0,4.2) {$G$};
  \draw[black] (n0) -- (n1) node[midway, right, font=\scriptsize]{$\frac{N_1}{N_0}$ simple};
  \draw[black] (n1) -- (n2) node[midway, right, font=\scriptsize]{$\frac{N_2}{N_1}$ simple};
  \draw[black] (dots) -- (nk);
  \node[font=\scriptsize, black, anchor=west] at (0.9,3.3) {each factor simple};
\end{tikzpicture}
$$

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

> **Theorem (Jordan-Hölder).** Every finite group $G \neq 1$ has a composition
> series, and any two composition series have the same length and the same
> composition factors up to isomorphism and reordering: if $1 = N_0 \trianglelefteq
> \cdots \trianglelefteq N_r = G$ and $1 = M_0 \trianglelefteq \cdots \trianglelefteq M_s = G$
> are two, then $r = s$ and some permutation matches
> $N_{i+1}/N_i \cong M_{\pi(i)+1}/M_{\pi(i)}$.

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

$$
% caption: Two composition series for $D_8$: the chains differ, but both produce
% the multiset of factors $\{\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_2\}$.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  nd/.style={draw, inner sep=2.5pt, fill=acc!8, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  % series 1
  \node[nd] (a0) at (0,0)   {$1$};
  \node[nd] (a1) at (0,1.1) {$< s >$};
  \node[nd] (a2) at (0,2.2) {$< s$, $r^2 >$};
  \node[nd, draw=acc] (a3) at (0,3.3) {$D_8$};
  \foreach \x/\y in {a0/a1, a1/a2, a2/a3}
    \draw[black] (\x) -- (\y) node[midway, left, font=\scriptsize]{$\mathbb{Z}_2$};
  % series 2
  \begin{scope}[xshift=4.0cm]
    \node[nd] (b0) at (0,0)   {$1$};
    \node[nd] (b1) at (0,1.1) {$< r^2 >$};
    \node[nd] (b2) at (0,2.2) {$< r >$};
    \node[nd, draw=acc] (b3) at (0,3.3) {$D_8$};
    \foreach \x/\y in {b0/b1, b1/b2, b2/b3}
      \draw[black] (\x) -- (\y) node[midway, right, font=\scriptsize]{$\mathbb{Z}_2$};
  \end{scope}
  \node[black, font=\scriptsize] at (2.0,-0.7) {same factors, distinct chains};
\end{tikzpicture}
$$

Counting all composition series is a lattice exercise. In $Q_8$ every maximal
chain descends through one of the three order-$4$ subgroups
$\langle i \rangle, \langle j \rangle, \langle k \rangle$, 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
$\{\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_2\}$.[^df-count] 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.[^df-holder]
- **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](/abstract-algebra/galois-theory/solvability-by-radicals-and-the-quintic).

> **Definition (Solvable group).** $G$ is **solvable** if it has a chain
> $1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G$
> with every quotient $G_{i+1}/G_i$ abelian.

A finite group is solvable exactly when all of its composition factors have
prime order.[^df-solv] 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](/abstract-algebra/subgroups-and-quotients/isomorphism-theorems)
directly. Take an abelian-quotient chain
$1 = N_0 \trianglelefteq \cdots \trianglelefteq N_n = N$ for $N$ and one for the
quotient, $1 = \overline{G}_0 \trianglelefteq \cdots \trianglelefteq \overline{G}_m = G/N$.
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:

$$
G_{i+1}/G_i \;\cong\; \bigl(G_{i+1}/N\bigr) \big/ \bigl(G_i/N\bigr)
= \overline{G}_{i+1}/\overline{G}_i,
$$

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](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups).
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 $S_n = \langle T \rangle$ 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

$$
\Delta = \prod_{1 \le i < j \le n} (x_i - x_j),
\qquad
\sigma(\Delta) = \prod_{1 \le i < j \le n} (x_{\sigma(i)} - x_{\sigma(j)}).
$$

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

> **Theorem (Sign homomorphism).** The map $\varepsilon : S_n \to \{\pm 1\}$ is
> a surjective homomorphism onto the multiplicative group of order two, and
> every transposition has sign $-1$.

That $\varepsilon(\tau\sigma) = \varepsilon(\tau)\varepsilon(\sigma)$ follows by
tracking sign changes when $\tau$ permutes the factors of $\sigma(\Delta)$;
computing $\varepsilon$ on $(1\,2)$ flips exactly one factor, and any
transposition is conjugate to $(1\,2)$, so all are odd.[^df-sign]

> **Definition (Alternating group).** The **alternating group** $A_n$ is the
> kernel of $\varepsilon$, the set of **even** permutations. Its complement is
> the set of **odd** permutations.

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

$$
% caption: The sign homomorphism partitions $S_n$ into the even permutations
% (the kernel $A_n$) and the odd permutations, mapping onto the group of order
% two.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  bl/.style={draw, minimum width=26mm, minimum height=11mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[bl, fill=acc!12, draw=acc] (ev) at (0,0.8)  {$A_n$\\(even, the kernel)};
  \node[bl] (od) at (0,-0.8) {odd coset};
  \node[bl] (c) at (5.2,0) {$C_2$ (order two)};
  \draw[->, acc, thick] (ev) -- (c) node[midway, above, font=\scriptsize]{even, sign $+1$};
  \draw[->, black, thick] (od) -- (c) node[midway, below, font=\scriptsize]{odd, other sign};
  \draw[black] (-1.7,-1.5) -- (-1.7,1.5);
  \draw[black] (-1.7,1.5) -- (-1.55,1.5); \draw[black] (-1.7,-1.5) -- (-1.55,-1.5);
  \node[black, font=\scriptsize, anchor=east] at (-1.8,0) {$S_n$};
\end{tikzpicture}
$$

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 $\alpha_1 \cdots \alpha_k$,
$\varepsilon(\sigma) = \prod \varepsilon(\alpha_i)$, giving a clean rule.[^df-cyc]

> **Proposition.** A permutation $\sigma$ is odd if and only if the number of
> even-length cycles in its cycle decomposition is odd.

> **Worked example.** Read the sign of a permutation from its cycle type. Take
> $\sigma = (1\,12\,8\,10\,4)(2\,13)(5\,11\,7)(6\,9) \in S_{13}$. The cycle
> lengths are $5, 2, 3, 2$; the even-length cycles number two, an even count, so
> $\sigma$ is **even** despite containing two transpositions. As a check, the
> transposition count is $(5-1) + (2-1) + (3-1) + (2-1) = 8$, even. By contrast
> $(1\,2\,3\,4\,5\,6)(7\,8)(9\,10\,11\,12)$ has cycle lengths $6, 2, 4$ (three
> even-length cycles, an odd count), so it is odd. Parity is not the parity of
> the order: $(1\,2)$ and $(1\,2)(3\,4)$ both have order $2$, yet one is odd and
> the other even; and any $\sigma$ of odd order is an even permutation, since all
> its cycle lengths are odd.

## The alternating group and simplicity

For small $n$ the alternating groups are familiar: $A_1$ and $A_2$ are trivial,
$A_3 = \langle (1\,2\,3) \rangle \cong \mathbb{Z}/3\mathbb{Z}$, 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.

$$
% caption: The subgroup lattice of $A_4$; the normal $V_4$ sits below $A_4$, the
% four order-three subgroups attach directly to $A_4$, and there is no subgroup
% of order six.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  nd/.style={draw, inner sep=2pt, fill=acc!8, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[nd, draw=acc] (G) at (0,3.4) {$A_4$};
  \node[nd] (V) at (-3.2,1.9) {$V_4$};
  \node[nd] (t1) at (-0.6,1.9) {$< (1\,2\,3) >$};
  \node[nd] (t2) at (1.2,1.9)  {$< (1\,2\,4) >$};
  \node[nd] (t3) at (3.0,1.9)  {$< (1\,3\,4) >$};
  \node[nd] (t4) at (4.8,1.9)  {$< (2\,3\,4) >$};
  \node[nd] (v1) at (-5.0,0.4) {$< (12)(34) >$};
  \node[nd] (v2) at (-3.2,0.4) {$< (13)(24) >$};
  \node[nd] (v3) at (-1.4,0.4) {$< (14)(23) >$};
  \node[nd] (one) at (0,-1.1) {$1$};
  \draw[black] (V) -- (G);
  \draw[black] (t1) -- (G); \draw[black] (t2) -- (G);
  \draw[black] (t3) -- (G); \draw[black] (t4) -- (G);
  \draw[black] (v1) -- (V); \draw[black] (v2) -- (V); \draw[black] (v3) -- (V);
  \draw[black] (one) -- (v1); \draw[black] (one) -- (v2); \draw[black] (one) -- (v3);
  \draw[black] (one) to[out=20,in=-90] (t1);
  \draw[black] (one) to[out=30,in=-90] (t2);
  \draw[black] (one) to[out=40,in=-90] (t3);
  \draw[black] (one) to[out=50,in=-90] (t4);
\end{tikzpicture}
$$

The normal $V_4$ makes $A_4$ solvable, with composition series
$1 \trianglelefteq \langle (1\,2)(3\,4) \rangle \trianglelefteq V_4 \trianglelefteq A_4$
and factors $\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_3$, all of prime order. From
$n = 5$ on, the alternating groups are simple.

> **Theorem.** $A_n$ is a non-abelian simple group for every $n \ge 5$.

The proof, deferred to the study of
[group actions](/abstract-algebra/group-actions-and-sylow/automorphisms-and-simple-groups),
turns on the $3$-cycles: they generate $A_n$, and a normal subgroup containing
one $3$-cycle must contain all of them.[^ju-simple]
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.

[^df-cauchy]: **Dummit & Foote**, _Abstract Algebra_, §3.4, Proposition 21 — the abelian case of Cauchy's theorem by induction on $N$ and $G/N$.
[^df-simple]: **Dummit & Foote**, §3.4 — the definition of a simple group; abelian simple groups are the $\mathbb{Z}/p\mathbb{Z}$, and the smallest non-abelian simple group has order $60$.
[^df-jh]: **Dummit & Foote**, §3.4, Theorem 22 (Jordan-Hölder) — existence of a composition series and uniqueness of the composition factors up to order and isomorphism.
[^df-count]: **Dummit & Foote**, §3.4, Exercise 2 — the 3 composition series of $Q_8$ and the 7 of $D_8$, with their composition factors.
[^df-holder]: **Dummit & Foote**, §3.4 — the Hölder program, the classification of finite simple groups (18 families and 26 sporadic groups), and the extension problem.
[^df-solv]: **Dummit & Foote**, §3.4 — solvable groups, the equivalence with prime-order composition factors, and closure under subgroups, quotients, and extensions.
[^df-sign]: **Dummit & Foote**, §3.5, Propositions 23–24 — the sign homomorphism $\varepsilon : S_n \to \{\pm 1\}$ built from the polynomial $\Delta$, and that transpositions are odd.
[^df-cyc]: **Dummit & Foote**, §3.5, Proposition 25 — the sign of a permutation from the parity of the number of even-length cycles.
[^ju-simple]: **Judson**, _Abstract Algebra: Theory and Applications_, §10.2 — the simplicity of $A_n$ for $n \ge 5$ via generation by $3$-cycles.