Automorphism Groups of Some Groups

Proposition

If $p$ is an odd prime, then $\mathrm{Aut}\left(Z_{p^n}\right)\cong Z_{\phi \left(p^n\right)}$ with $n⩾1$. If $p=2$, then $\mathrm{Aut}\left(Z_{2^n}\right)\cong Z_2\times Z_{2^{n-2}}$ with $n\ge 2$.

\begin{proof} The case of odd prime is easy, using the existence of primitive root.

For the case of $2^n$, we can prove that $a^{2^{k-2}}\equiv 1(\mathrm{mod}{2}^k)$ and $3^{2^{k-3}}\not\equiv 1\mathrm{mod}{2}^k$ with $k⩾3$ by induction. Then we finish the proof by here. \end{proof}

Proposition

Let $D_{2n}$ be a dihedral group. Then:

• $\mathrm{Aut}(D_4)\cong S_3$;
• $\mathrm{Aut}(D_{2n})\cong {\mathbb{Z}}_n:\mathrm{Aut}({\mathbb{Z}}_n)$ when $n\ge 3$.

\begin{proof} When $n=2$, the group $D_4\cong {\mathbb{Z}}_2\times {\mathbb{Z}}_2$. Suppose $D_4=\{1,a,b,ab\}$ where $|a|=|b|=|ab|=2$. For any $\phi \in \mathrm{Aut}(D_4)$, there is $\phi (1)=1$. Thus $|\mathrm{Aut}(D_4)|\le 3!=6$. Furthermore, since any element of $\mathrm{Sym}\{a,b,ab\}$ is an automorphism, we have $\mathrm{Aut}(D_4)\cong S_3$.

When $n\ge 3$, define $D_{2n}$ as

$$D_{2n}=⟨x,y:|x|=n,|y|=2,x^y=x^{-1}⟩.$$

Then for any $\phi \in \mathrm{Aut}(D_{2n})$, it satisfies $|\phi (x)|=n$ and $|\phi (y)|=2$. Hence

$$\begin{array}{rl}& \phi (x)\in \{x^l:(l,n)=1\},\\ & \phi (y)\in \{x^{m}y:0\le m\le n-1\}.\end{array}$$

Let $A:=⟨{\phi }_m:0\le m\le n-1⟩$ and $B:=⟨{\phi }^l:(l,n)=1⟩$, where ${\phi }_m(x)=x,{\phi }_m(y)=x^{m}y$ and ${\phi }^l(x)=x^l,{\phi }^l(y)=y$. Then $\mathrm{Aut}(D_{2n})=⟨A,B⟩$ and $A\cap B=1$. It is easy to verify that $B⊲\mathrm{Aut}(D_{2n})$. Therefore, we have

$$\mathrm{Aut}(D_{2n})=A:B\cong {\mathbb{Z}}_n:\mathrm{Aut}({\mathbb{Z}}_n)$$

and so the proof is completed. \end{proof}

Proposition

$\mathrm{Aut}(Q_8)=S_4$.

\begin{proof} Note that $Q_8$ has three cyclic subgroups of order $4$: $⟨i⟩,⟨j⟩,⟨k⟩$, and $\mathrm{Aut}\left(Q_8\right)$ acts on these three subgroups; inducing a homomorphism $\Phi :\mathrm{Aut}\left(Q_8\right)\to S_3$. We can see that, the homomorphism is surjective, since the two automorphisms $f:i\mapsto j,j\mapsto i$, and $g:j\mapsto k,k\mapsto j$ give two transpositions in $S_3$. Thus $\Phi (\mathrm{Aut}(Q_8))\cong S_3$.

The kernel contains those $\phi \in \mathrm{Aut}(G)$ such that $\phi (⟨i⟩)=⟨i⟩$, $\phi (⟨j⟩)=⟨j⟩$ and $\phi (⟨k⟩)=⟨k⟩$. Notice that $\phi (⟨i⟩)=⟨i⟩$ means $\phi (i)\in \{i,-i\}$, and similarly, $\phi (j)\in \{j,-j\}$. One can check that these four choices are automorphisms of order $2$ (or $1$) (since they are switching elements in a pair), and hence the kernel is Klein-4 group $V_4$.

Consider the following automorphisms: $S:i\mapsto j,j\mapsto i$ and $T:j\mapsto k,k\mapsto j$. Since $S$ and $T$ fix subgroup $⟨k⟩$ and $⟨i⟩$ respectively, they are not inner. Therefore, we have $⟨S,T⟩=K⩽\mathrm{Aut}\left(Q_8\right)$, such that $K\cong S_3$ and $\Phi (K)=S_3$. Also,

$$\ker (\Phi )\cap K=1$$

Therefore, $\mathrm{Aut}\left(Q_8\right)=\ker (\Phi )⋊K\cong V_4⋊S_3$.

Consider an element $f:i\mapsto -i,j\mapsto j$ of $\ker (\Phi )$, and two elements of $K\cong \mathrm{Im}$: $g:i\mapsto j,j\mapsto i$ (like a transposition), and $h:i\mapsto j,j\mapsto k$ (like a 3-cycle). One can check that $f$ doesn’t commute with $g$ as well as $h$. In fact, this shows that no element of $V_4\backslash \{1\}$ commutes with any element of $K\backslash \{1\}$. This means, the action of $K$ on $V_4$ (by conjugation) is faithful; and up to equivalence, there is only one such action. Therefore, $\mathrm{Aut}\left(Q_8\right)=V_4⋊K\cong S_4$. \end{proof}

Proposition

The following holds:

• $\mathrm{Aut}(p_{-}^{1+2})\cong p_{-}^{1+2}:{\mathbb{Z}}_{p-1}$;
• $\mathrm{Aut}(p_{+}^{1+2})\cong {\mathrm{GL}}_2(p)$.

\begin{proof} i) todo Assume that $G:=p_{-}^{1+2}=⟨a⟩:⟨b⟩$ with $|a|=p^2$ and $|b|=p$. Then $G/\Phi (G)\cong {\mathbb{Z}}_p^2$ and $\Phi (G)=Z(G)=G^{\prime }=\{1,a^p,\cdots ,a^{p(p-1)}\}$. For any nontrivial $\phi \in \mathrm{Aut}(G)$, $\phi$ is a nontrivial automorphism of $G/\Phi (G)$. Thus, $\mathrm{Aut}(G)\lesssim {\mathrm{GL}}_2(p)$. Note that $[a,b]=a^{-1}{a}^b=a^{kp}$ for some $k\in \mathbb{Z}$, we may suppose that $a^b=a^{kp+1}$.

(to be continued)

ii) Since $G=⟨a,b|a^p=b^p=c^p=1,[a,b]=c⟩$ by the argument of ^oymbgb, it is easy to prove.

\end{proof}