Let $G_1,G_2$ be groups. $G_1 \cong G_2 \implies\operatorname{Aut}(G_1) \cong \operatorname{Aut}(G_2)$.

automorphism-groupgroup-isomorphismgroup-theorysolution-verification

Let $G_1,G_2$ be groups.
$$
G_1 \cong G_2 \implies \operatorname{Aut}(G_1) \cong \operatorname{Aut}(G_2).
$$

My solution :

Since $G_1 \cong G_2$ there is an isomorphism $f : G_1 \to G_2$.

Denote $\phi:\operatorname{Aut}(G_1)\to \operatorname{Aut(G_2)}$ such that $\phi
(\varphi)=f\varphi f^{-1}$

$\phi$ is an homomorphism:

$\phi(\varphi_1\varphi_2)=f\varphi_1\varphi_2f^{-1}=f\varphi_1f^{-1}f\varphi_2f^{-1}=\phi(\varphi_1)\phi(\varphi_2)$

$\phi$ is injective:

Suppose $\phi$ isn't injective $\implies$ exist $\varphi_1\neq\varphi_2\space \text{ such that }
\phi(\varphi_1)=\phi(\varphi_2) $

$\phi(\varphi_1)=\phi(\varphi_2)\implies f\varphi_1f^{-1}=f\varphi_2f^{-1}\implies \varphi_1=\varphi_2 , Contradiction.$

How am I supposed to prove that $\phi$ is surjective ?

I got a bit confused.

Best Answer

Now that you've worked out the most, let me add some details (and the surjectivity part). For $f\colon G_1\longrightarrow G_2$ isomorphism (which exists by assumption), your candidate isomorphism is: \begin{alignat*}{2} \phi:\operatorname{Aut}(G_1)&\longrightarrow&\operatorname{Aut}(G_2) \\ \varphi&\longmapsto& \phi_\varphi: G_2 &\longrightarrow G_2 \\ &&g&\longmapsto \phi_\varphi(g):=(f\varphi f^{-1})(g) \\ \tag 1 \end{alignat*} First, you need to prove that $\phi$ is well-defined, which in this case means that $\phi_\varphi$ is indeed an automorphism of $G_2$, for every $\varphi\in\operatorname{Aut}(G_1)$. In fact:

  1. for every $\varphi\in\operatorname{Aut}(G_1)$, $\phi_\varphi=f\varphi f^{-1}$ is a composition of (right to left): a bijection from $G_2$ to $G_1$, a bijection from $G_1$ to itself, and a bijection from $G_1$ to $G_2$; therefore, $\phi_\varphi$ is a bijection from $G_2$ to itself;
  2. since $f,\varphi,f^{-1}$ are all operation-preserving maps, then, for every $\varphi\in\operatorname{Aut}(G_1)$ and $g,h\in G_2$: $\phi_\varphi(gh)=(f\varphi f^{-1})(gh)=$ $f(\varphi(f^{-1}(gh))=$ $f(\varphi(f^{-1}(g)f^{-1}(h)))=$ $f(\varphi(f^{-1}(g))\varphi(f^{-1}(h)))=$ $f(\varphi(f^{-1}(g)))f(\varphi(f^{-1}(h)))=$ $(f\varphi f^{-1})(g)(f\varphi f^{-1})(h)=$ $\phi_\varphi(g)\phi_\varphi(h)$.

Therefore, indeed $\phi_\varphi\in\operatorname{Aut}(G_2)$ for every $\varphi\in\operatorname{Aut}(G_1)$. This proves that $\phi$ in $(1)$ is well-defined.

Injectivity of $\phi$ (here you can avoid the contradiction argument): $\phi_\varphi=\phi_{\varphi'}\Longrightarrow$ $f\varphi f^{-1}=$ $f\varphi' f^{-1}\Longrightarrow$ $\varphi=\varphi'$.

Surjectivity of $\phi$: for every $\psi\in\operatorname{Aut}(G_2)$, we have that $f^{-1}\psi f\in\operatorname{Aut}(G_1)$ (the proof mimics exactly the one about the good-definiteness of $\phi$ above) and $\psi=$ $(ff^{-1})\psi (ff^{-1})=$ $f(f^{-1}\psi f)f^{-1}=$ $\varphi_{f^{-1}\psi f}$.

You have already shown that $\phi$ preserves the operation (composition).

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