---
title: Automorphisms and Simplicity of Aₙ
module: Group Actions and Sylow Theory
moduleNumber: 4
lessonNumber: 4
order: 404
summary: >
  Conjugation makes a group act on itself and on its normal subgroups by
  automorphisms, giving the inner automorphism group G/Z(G) and the embedding of
  N(H)/C(H) into Aut(H). Characteristic subgroups are those every automorphism
  fixes, and the automorphism group of a cyclic group is its unit group. The
  lesson closes by proving the alternating group Aₙ is simple for n ≥ 5.
topics: [Group Actions and Sylow Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 4 Group Actions; §4.4 Automorphisms, §4.6 The Simplicity of A_n"
draft: false
---

Conjugation appeared in the [class equation](/abstract-algebra/group-actions-and-sylow/conjugation-and-the-class-equation)
as an action of $G$ on its own elements. Reinterpreted, each conjugation is a
_structure-preserving_ bijection — an automorphism — so the conjugation action is
really a homomorphism from $G$ into a group of symmetries of $G$. This yields the
inner automorphism group, the notion of a characteristic subgroup, tools to
compute $\mathrm{Aut}(G)$ for cyclic groups, and a proof that $A_n$ is simple for
all $n \ge 5$.[^df-aut]

## The automorphism group

> **Definition (Automorphism group).** An **automorphism** of $G$ is an isomorphism
> $G \to G$. The set $\mathrm{Aut}(G)$ of all automorphisms is a group under
> composition, a subgroup of the symmetric group $S_G$.

Automorphisms are permutations of the underlying set that also respect the
operation, so $\mathrm{Aut}(G) \le S_G$. The richest supply comes from
conjugation, and it works not only on $G$ but on any normal subgroup.[^df-prop13]

> **Proposition (Conjugation acts by automorphisms).** Let $H \trianglelefteq G$.
> For each $g \in G$ the map $\varphi_g : h \mapsto ghg^{-1}$ is an automorphism of
> $H$, and $g \mapsto \varphi_g$ is a homomorphism $G \to \mathrm{Aut}(H)$ with
> kernel $C_G(H)$. Hence $G / C_G(H)$ embeds in $\mathrm{Aut}(H)$.

Normality makes $\varphi_g$ map $H$ into $H$; it is a homomorphism because
$\varphi_g(hk) = ghkg^{-1} = (ghg^{-1})(gkg^{-1}) = \varphi_g(h)\varphi_g(k)$, and
bijective with inverse $\varphi_{g^{-1}}$. An element acts trivially exactly when
it commutes with every $h \in H$, so the kernel is the
[centralizer](/abstract-algebra/subgroups-and-quotients/subgroups-and-substructures)
$C_G(H)$; the first isomorphism theorem gives the embedding. Two corollaries fall
out at once.[^df-cor15]

- **Conjugate subgroups are isomorphic.** For any $K \le G$ and $g \in G$,
  conjugation by $g$ carries $K$ isomorphically onto $gKg^{-1}$; conjugate elements
  and subgroups share order.
- **The $N/C$ theorem.** For any $H \le G$, the quotient $N_G(H) / C_G(H)$ embeds
  in $\mathrm{Aut}(H)$. Taking $H = G$ gives $G / Z(G) \hookrightarrow
  \mathrm{Aut}(G)$.

### Inner automorphisms

Conjugations form a distinguished piece of the automorphism group.

> **Definition (Inner automorphism).** Conjugation by a fixed $g \in G$ is an
> **inner automorphism**. The inner automorphisms form a subgroup
> $\mathrm{Inn}(G) \le \mathrm{Aut}(G)$, and $\mathrm{Inn}(G) \cong G / Z(G)$.

The isomorphism is the $H = G$ case of the proposition: the map $g \mapsto
\varphi_g$ has image $\mathrm{Inn}(G)$ and kernel $Z(G)$. So a group with trivial
center is isomorphic to its own inner automorphism group, while an abelian group
has only the trivial inner automorphism. The inner automorphisms are the group's
symmetries realized as a rearrangement of the Cayley table by
conjugation.

$$
% caption: An inner automorphism $\varphi_g$ relabels every element by $x\mapsto
% gxg^{-1}$; it is trivial exactly on the center, so $\mathrm{Inn}(G)\cong G/Z(G)$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw=black, minimum width=20mm, minimum height=10mm] (G) at (0,0) {$G$};
\node[draw=black, minimum width=20mm, minimum height=10mm] (G2) at (5.2,0) {$G$};
\draw[->, acc, thick] (G) -- (G2) node[midway, above, font=\footnotesize] {conjugate by $g$};
\node[font=\footnotesize, text=black, anchor=north] at (2.6,-0.65) {each element sent to its conjugate};
% quotient row
\node[draw=acc, fill=acc!12, minimum width=26mm, minimum height=9mm] (quot) at (2.6,-2.4) {$\mathrm{Inn}(G)$};
\draw[->, acc] (G2.south) .. controls (5.2,-1.6) and (4.0,-2.4) .. (quot.east);
\node[font=\footnotesize, text=black, anchor=west] at (4.6,-2.4) {center acts trivially};
\end{tikzpicture}
$$

## Characteristic subgroups

Normality means _stable under inner automorphisms_. Strengthening this to _all_
automorphisms gives a stronger notion.[^df-char]

> **Definition (Characteristic subgroup).** $H \le G$ is **characteristic**,
> written $H \ \mathrm{char}\ G$, if $\sigma(H) = H$ for every $\sigma \in
> \mathrm{Aut}(G)$.

Since inner automorphisms are a subset of all automorphisms, characteristic is
stronger than normal. Three facts describe how characteristic subgroups sit inside
larger groups.

- **Characteristic implies normal.** Every inner automorphism fixes $H$
  setwise, so $H \trianglelefteq G$.
- **Unique order implies characteristic.** If $H$ is the only subgroup of $G$ of
  its order, any automorphism sends $H$ to a subgroup of that same order, hence
  back to $H$.
- **Transitivity across normality.** If $K \ \mathrm{char}\ H$ and $H
  \trianglelefteq G$, then $K \trianglelefteq G$. Normality alone is _not_
  transitive, but a characteristic subgroup of a normal subgroup is normal.

$$
% caption: Containment of the three conditions: characteristic subgroups lie
% inside the normal ones, which lie inside all subgroups; each inclusion is strict.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% nested boxes: characteristic inside normal inside all subgroups
\draw[draw=black, fill=black!4] (0,0) rectangle (9,4.4);
\node[font=\footnotesize, text=black] at (4.5,4.05) {all subgroups};
\draw[draw=black, fill=acc!7] (0.5,0.35) rectangle (6.5,3.5);
\node[font=\footnotesize, text=black] at (3.5,3.15) {normal};
\draw[draw=acc, fill=acc!16] (1.0,0.7) rectangle (4.6,2.35);
\node[font=\footnotesize, text=acc] at (2.8,1.5) {characteristic};
% legend below the box
\node[font=\footnotesize, text=black, anchor=west] at (0.2,-0.55) {preserved by inner autos: normal};
\node[font=\footnotesize, text=acc, anchor=west] at (0.2,-1.05) {preserved by all autos: characteristic};
\end{tikzpicture}
$$

## Automorphisms of cyclic groups

A cyclic group's automorphisms are determined by where they send a generator, and
they form the [unit group](/abstract-algebra/subgroups-and-quotients/cyclic-groups)
modulo $n$.[^df-prop16]

> **Proposition ($\mathrm{Aut}$ of a cyclic group).** The automorphism group of the
> cyclic group of order $n$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times$, an
> abelian group of order $\varphi(n)$.

Let $x$ generate $\mathbb{Z}_n$. An automorphism $\psi$ is determined by $\psi(x) =
x^a$, and $\psi$ preserves order only if $\gcd(a, n) = 1$; conversely every such
$a$ gives an automorphism. The map $\psi \mapsto a \bmod n$ is an isomorphism
$\mathrm{Aut}(\mathbb{Z}_n) \to (\mathbb{Z}/n\mathbb{Z})^\times$, because composing
$x \mapsto x^a$ with $x \mapsto x^b$ multiplies the exponents.

$$
% caption: $\mathrm{Aut}(\mathbb{Z}_8)\cong(\mathbb{Z}/8\mathbb{Z})^\times$: each
% unit $a$ names the automorphism sending a generator $x$ to $x^a$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[font=\footnotesize, text=black] at (0,2.3) {unit $a$ mod $8$};
\node[font=\footnotesize, text=black] at (4.6,2.3) {image of $x$};
\foreach \i/\a in {0/1, 1/3, 2/5, 3/7} {
  \node[draw=acc, fill=acc!12, minimum width=8mm, minimum height=6mm] (u\i) at (0,-\i*0.85+1.5) {$\a$};
  \node[draw=black, minimum width=18mm, minimum height=6mm] (a\i) at (4.6,-\i*0.85+1.5) {$x^{\a}$};
  \draw[->, acc] (u\i.east) -- (a\i.west);
}
\node[font=\footnotesize, text=black, anchor=north] at (2.3,-1.9) {count $= 4$ units};
\end{tikzpicture}
$$

> **Worked example.** Compute $\mathrm{Aut}(\mathbb{Z}_{10})$. The units modulo
> $10$ are $\{1, 3, 7, 9\}$, so $|\mathrm{Aut}(\mathbb{Z}_{10})| = \varphi(10) = 4$.
> The unit $3$ has powers $3^1 = 3$, $3^2 = 9$, $3^3 \equiv 7$, $3^4 \equiv 1
> \pmod{10}$, running through all four units, so $(\mathbb{Z}/10\mathbb{Z})^\times =
> \langle 3 \rangle$ is cyclic of order $4$. Hence $\mathrm{Aut}(\mathbb{Z}_{10})
> \cong \mathbb{Z}_4$, generated by the automorphism $x \mapsto x^3$.[^df-prop16]

A handful of small automorphism groups recur throughout the subject:[^df-prop17]

| Group $G$ | $\mathrm{Aut}(G)$ | Order | Note |
| --- | --- | --- | --- |
| $\mathbb{Z}_n$ | $(\mathbb{Z}/n\mathbb{Z})^\times$ | $\varphi(n)$ | abelian |
| $\mathbb{Z}_p$ (prime) | $\mathbb{Z}_{p-1}$ | $p-1$ | cyclic |
| $V_4 = \mathbb{Z}_2 \times \mathbb{Z}_2$ | $S_3$ | $6$ | permutes the $3$ involutions |
| $Q_8$ | $S_4$ | $24$ | — |
| $D_8$ | $D_8$ | $8$ | — |
| $S_n$, $n \ne 6$ | $S_n$ | $n!$ | all automorphisms inner |

The elementary abelian group $V = (\mathbb{Z}/p)^n$ is a vector space over
$\mathbb{F}_p$, and its automorphisms are the invertible linear maps, so
$\mathrm{Aut}(V) \cong \mathrm{GL}_n(\mathbb{F}_p)$; the case $V_4 \cong
(\mathbb{Z}/2)^2$ gives $\mathrm{GL}_2(\mathbb{F}_2) \cong S_3$. These computations
feed the classification of [semidirect products](/abstract-algebra/products-and-group-structure/semidirect-products),
where a group is assembled from a normal subgroup $H$ and an action into
$\mathrm{Aut}(H)$.

### Worked example: groups of order pq

The $N/C$ theorem, applied through $\mathrm{Aut}(\mathbb{Z}_q)$, forces some groups
to be abelian outright. Let $|G| = pq$ with primes $p \le q$ and $p \nmid q - 1$;
then $G$ is cyclic.[^df-order-pq]

By the [Sylow count](/abstract-algebra/group-actions-and-sylow/sylow-theorems),
$n_q = 1$, so the Sylow $q$-subgroup $H = \langle x \rangle$ is normal. The $N/C$
theorem gives $G / C_G(H) \hookrightarrow \mathrm{Aut}(H) \cong
(\mathbb{Z}/q\mathbb{Z})^\times$, a group of order $q - 1$. Since $H$ is abelian,
$H \le C_G(H)$, so $q \mid |C_G(H)|$ and $[G : C_G(H)]$ divides both $p$ and $q -
1$. The hypothesis $p \nmid q - 1$ leaves $[G : C_G(H)] = 1$, i.e. $C_G(H) = G$, so
$H \le Z(G)$. Then $G / Z(G)$ has order dividing $p$, hence is cyclic, which forces
$G$ abelian; a product of an element of order $q$ with one of order $p$ then has
order $pq$, so $G \cong \mathbb{Z}_{pq}$. When instead $p \mid q - 1$, the
nontrivial homomorphism $\mathbb{Z}_p \to \mathrm{Aut}(\mathbb{Z}_q)$ builds a
nonabelian
[semidirect product](/abstract-algebra/products-and-group-structure/semidirect-products),
and the automorphism group determines which of the two outcomes occurs.

> **Worked example.** Every group of order $15$ is cyclic. Write $15 = 3 \cdot 5$
> with $p = 3$, $q = 5$; the condition $p \nmid q - 1$ holds because $3 \nmid 4$.
> The Sylow $5$-subgroup $H$ is normal, and $G / C_G(H) \hookrightarrow
> \mathrm{Aut}(\mathbb{Z}_5) \cong (\mathbb{Z}/5\mathbb{Z})^\times$ of order $4$;
> since $[G : C_G(H)]$ divides both $3$ and $4$, it equals $1$, so $H \le Z(G)$.
> Then $G / Z(G)$ has order dividing $3$, hence is cyclic, forcing $G$ abelian, and
> an element of order $3$ times one of order $5$ has order $15$: $G \cong
> \mathbb{Z}_{15}$. Order $15$ is the smallest composite order carrying a single
> group.[^df-order-pq]

## The simplicity of Aₙ

A [simple group](/abstract-algebra/group-actions-and-sylow/sylow-theorems) has no
proper nontrivial normal subgroup. The alternating groups are the first infinite
family of nonabelian simple groups, and their simplicity is the group-theoretic
fact behind the unsolvability of the general quintic. Note that $A_3 \cong
\mathbb{Z}_3$ is simple while $A_4$ is _not_ (its Klein four-group is normal), so
the theorem starts at $n = 5$.[^df-simplicity]

> **Theorem (Simplicity of $A_n$).** $A_n$ is simple for all $n \ge 5$.

The base case $n = 5$ is settled by the class-equation count in the
[conjugation lesson](/abstract-algebra/group-actions-and-sylow/conjugation-and-the-class-equation):
the classes of $A_5$ have sizes $1, 15, 20, 12, 12$, and no sub-collection
including the identity sums to a proper divisor of $60$, so $A_5$ has no proper
nontrivial normal subgroup. The general case is induction on $n$.

Assume $n \ge 6$, set $G = A_n$, and suppose $H \trianglelefteq G$ with $H \ne 1,
G$. For each point $i$ let $G_i$ be the stabilizer of $i$ in the natural action on
$\{1, \dots, n\}$; each $G_i \cong A_{n-1}$ is simple by induction. The proof runs
in two moves.

**A fixed point pulls in a whole stabilizer.** If some nonidentity $\tau \in H$
fixes a point $i$, then $\tau \in H \cap G_i$, a normal subgroup of the simple
group $G_i$, forcing $H \cap G_i = G_i$, so $G_i \le H$. Conjugating, $G_j \le H$
for every $j$. Any element of $A_n$ is a product of an even number of
transpositions, each pair of which lies in some $G_j$, so $\langle G_1, \dots, G_n
\rangle = A_n \le H$ — contradicting $H \ne G$.

$$
% caption: One fixed point forces a stabilizer $G_i\cong A_{n-1}$ into $H$;
% conjugates give every $G_j$, and together they generate all of $A_n$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\foreach \i/\x in {1/0, 2/2.0, 3/4.0} {
  \node[draw=acc, fill=acc!12, minimum width=15mm, minimum height=9mm] (g\i) at (\x,0) {$G_{\i}$};
}
\node[font=\footnotesize] at (5.6,0) {$\cdots$};
\node[draw=acc, fill=acc!12, minimum width=13mm, minimum height=9mm] (gn) at (7.1,0) {$G_{n}$};
\node[draw=black, fill=acc!5, minimum width=20mm, minimum height=9mm] (An) at (3.5,-2.0) {$A_n \le H$};
\foreach \g in {g1,g2,g3,gn} \draw[->, acc] (\g.south) -- (An.north);
\node[font=\footnotesize, text=black, anchor=west] at (-1.7,-2.0) {stabilizers generate};
\end{tikzpicture}
$$

**No nonidentity element may fix a point.** The previous paragraph shows a
nonidentity element of $H$ can fix nothing, so if $\tau_1, \tau_2 \in H$ agree at
even one point then $\tau_2^{-1}\tau_1$ fixes that point and must be the identity:
elements of $H$ are determined by their value at a single point. Now suppose some
$\tau \in H$ has a cycle of length $\ge 3$, say $\tau = (a_1\, a_2\, a_3 \cdots)
\cdots$. Choose $\alpha \in A_n$ fixing $a_1, a_2$ but moving $a_3$ (possible since
$n \ge 5$). Then $\tau_1 = \alpha\tau\alpha^{-1} \in H$ agrees with $\tau$ at $a_1$
yet differs from it, contradicting uniqueness. So every nonidentity element of $H$
is a product of disjoint $2$-cycles. Applying the same conjugation trick to $\tau =
(a_1\,a_2)(a_3\,a_4)\cdots$ with $\alpha = (a_1\,a_2)(a_3\,a_5)$ (using $n \ge 6$)
again produces a distinct element of $H$ agreeing with $\tau$ at $a_1$. Both cases
are impossible, so no such $H$ exists and $A_n$ is simple.[^df-simplicity]

The conjugation trick is the same relabeling principle from the class equation:
$\alpha\tau\alpha^{-1}$ has the cycle structure of $\tau$ with entries renamed by
$\alpha$. Choosing $\alpha$ to hold one entry and move another manufactures a
second element of $H$, impossible in a simple group. Simplicity of $A_5$
propagates up the whole family, and with it the fact that $S_n$ for $n \ge 5$ is
not [solvable](/abstract-algebra/products-and-group-structure/nilpotent-and-solvable-groups),
the obstruction that leaves the general quintic without a formula in radicals.

[^df-aut]: **Dummit & Foote**, §4.4 — Automorphisms: the automorphism group $\mathrm{Aut}(G)$ as a subgroup of $S_G$.
[^df-prop13]: **Dummit & Foote**, §4.4, Proposition 13: $G$ acts on a normal subgroup $H$ by automorphisms, giving $G/C_G(H) \hookrightarrow \mathrm{Aut}(H)$.
[^df-cor15]: **Dummit & Foote**, §4.4, Corollaries 14–15: conjugate subgroups are isomorphic, $N_G(H)/C_G(H) \hookrightarrow \mathrm{Aut}(H)$, and $\mathrm{Inn}(G) \cong G/Z(G)$.
[^df-char]: **Dummit & Foote**, §4.4 — characteristic subgroups: definition and the three properties (characteristic implies normal, unique-order subgroups are characteristic, characteristic-in-normal is normal).
[^df-prop16]: **Dummit & Foote**, §4.4, Proposition 16: $\mathrm{Aut}(\mathbb{Z}_n) \cong (\mathbb{Z}/n\mathbb{Z})^\times$ of order $\varphi(n)$.
[^df-prop17]: **Dummit & Foote**, §4.4, Proposition 17: automorphism groups of cyclic $p$-groups, elementary abelian groups ($\mathrm{GL}_n(\mathbb{F}_p)$), $S_n$, $D_8$, and $Q_8$.
[^df-order-pq]: **Dummit & Foote**, §4.4 — worked example following Proposition 16: a group of order $pq$ with $p \nmid q-1$ is cyclic, via $G/C_G(H) \hookrightarrow \mathrm{Aut}(\mathbb{Z}_q)$ of order $q-1$.
[^df-simplicity]: **Dummit & Foote**, §4.6, Theorem 24: $A_n$ is simple for $n \ge 5$, by induction using point stabilizers $G_i \cong A_{n-1}$ and the conjugation relabeling argument.