Automorphisms and Simplicity of Aₙ

Conjugation appeared in 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 $n\ge 5$.¹

The automorphism group

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.²

Normality makes ${\phi }_g$ map $H$ into $H$; it is a homomorphism because ${\phi }_g(hk)=ghk{g}^{-1}=(gh{g}^{-1})(gk{g}^{-1})={\phi }_g(h){\phi }_g(k)$, and bijective with inverse ${\phi }_{g^{-1}}$. An element acts trivially exactly when it commutes with every $h\in H$, so the kernel is the centralizer$C_G(H)$; the first isomorphism theorem gives the embedding. Two corollaries fall out at once.³

• Conjugate subgroups are isomorphic. For any $K\le G$ and $g\in G$, conjugation by $g$ carries $K$ isomorphically onto $gK{g}^{-1}$; 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.

The isomorphism is the $H=G$ case of the proposition: the map $g\mapsto {\phi }_g$ 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.

Characteristic subgroups

Normality means stable under inner automorphisms. Strengthening this to all automorphisms gives a stronger notion.⁴

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 $H⊴G$.
• 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 $H⊴G$, then $K⊴G$. Normality alone is not transitive, but a characteristic subgroup of a normal subgroup is normal.

Automorphisms of cyclic groups

A cyclic group's automorphisms are determined by where they send a generator, and they form the unit group modulo $n$.⁵

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 $\psi \mapsto a\mathrm{mod}n$ 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.

A handful of small automorphism groups recur throughout the subject:⁶

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, 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.⁷

By the Sylow count, $n_q=1$, so the Sylow $q$-subgroup $H=⟨x⟩$ 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, and the automorphism group determines which of the two outcomes occurs.

The simplicity of Aₙ

A simple group 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 $n=5$.⁸

The base case $n=5$ is settled by the class-equation count in the conjugation lesson: 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 $H⊴G$ 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 $⟨G_1,\dots ,G_n⟩=A_n\le H$ — contradicting $H\ne G$.

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.⁸

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, the obstruction that leaves the general quintic without a formula in radicals.