I'm trying to solve the following problem. I have a solution, I'm tripping over myself explaining part of it. Is my solution clear? (Particularly part 2) If not, can someone help me clarify it? Thanks!


If a group $G$ is isomorphic to $H$, prove that $\text{Aut}(G)$ is isomorphic to $\text{Aut}(H)$


Let $\mathcal{G} \in \text{Aut}(G)$ and $\phi:G \rightarrow H$. By assumption $\phi$ is an isomorphism. Define: $$\Phi:\text{Aut}(G) \rightarrow \text{Aut}(H) \\ \mathcal{G} \mapsto \phi\mathcal{G}\phi^{-1}$$We want to show $\Phi$ is an isomorphism.

Proceed in 3 steps: Show $\Phi$ injects, show $\Phi$ surjects, show $\Phi$ is a homomorphism.

  1. Suppose $\Phi(\mathcal{G})=\Phi(\mathcal{G'})$ $$\Phi(\mathcal{G})=\Phi(\mathcal{G'}) \\ \implies \phi\mathcal{G}\phi^{-1} = \phi\mathcal{G}'\phi^{-1} \\ \implies \mathcal{G} = \mathcal{G'} \text{ [by taking }\phi^{-1} \text{of both sides, followed by }\phi \text{]}$$

  2. Suppose we have $\mathcal{H} \in \text{Aut}(H)$. We want to show there exists $\mathcal{G}$ such that $\Phi(\mathcal{G})=\mathcal{H}$. Construct such a $\mathcal{G}$ by letting $\mathcal{G} = \phi^{-1}\mathcal{H}\phi$. To check $\phi^{-1}\mathcal{H}\phi \in \text{Aut}(G)$ we need to check $\phi^{-1}\mathcal{H}\phi$ is a bijection on $G$ and that its a homomorphism on $G$. Bijection of $\phi^{-1}\mathcal{H}\phi$ follows from the composition of 3 bijections. We check homomorphism by considering the action on 2 arbitrary elements of $G$ say $g_1$ and $g_2$. $$\phi^{-1}\mathcal{H}\phi(g_1 g_2)=\phi^{-1}\mathcal{H}[\phi(g_1)\phi(g_2)] = \phi^{-1}[\mathcal{H}\phi(g_1)\mathcal{H}\phi(g_2)]=\phi^{-1}\mathcal{H}\phi(g_1)\phi^{-1}\mathcal{H}\phi(g_2)$$ By homomorphic properties of $\phi$, $\phi^{-1}$, and $\mathcal{H}$. Thus $\phi^{-1}\mathcal{H}\phi \in \text{Aut}(G)$. We just need to check that the element of $\text{Aut}(G)$ so defined will send us to the given $\mathcal{H}$. $$\Phi(\phi^{-1}\mathcal{H}\phi) = \phi\phi^{-1}\mathcal{H}\phi\phi^{-1} = \mathcal{H}$$

  3. Let $\mathcal{G_1}$,$\mathcal{G_1} \in \text{Aut}(G)$. Then: $$\Phi(\mathcal{G_1}\mathcal{G_1})=\phi\mathcal{G_1}\mathcal{G_1}\phi^{-1}=\phi\mathcal{G_1}\phi^{-1}\phi\mathcal{G_1}\phi^{-1}=\Phi(\mathcal{G_1})\Phi(\mathcal{G_2})$$

  • $\begingroup$ Remember that the composition of two homomorphisms is always a homomorphism $\endgroup$ – leibnewtz Nov 18 '17 at 22:24
  • $\begingroup$ where did I use that and not realize it? $\endgroup$ – yoshi Nov 18 '17 at 22:28
  • $\begingroup$ I'm saying you don't use it when you should. In two places you check whether or not the maps you defined are homomorphisms, but you can just note that they're compositions $\endgroup$ – leibnewtz Nov 18 '17 at 22:33
  • $\begingroup$ aha, thanks for pointing this out. i would have gone on verifying this for the rest of my life, lol. $\endgroup$ – yoshi Nov 18 '17 at 22:47

It's correct, apart from one particular. It can also be simplified.

First, prove that, for $\alpha\in\operatorname{Aut}(G)$, $\phi\alpha\phi^{-1}\in\operatorname{Aut}(H)$: this is essentially obvious (but is missing from your proof). Thus $\Phi$ is well defined.

Second, prove that $\Psi\colon\operatorname{Aut}(H)\to\operatorname{Aut}(G)$ defined by $\Psi(\delta)=\phi^{-1}\delta\phi$ is the two-sided inverse of $\Phi$. Note there's no need to check $\Psi$ is well defined, because it's the same argument as above with $G$ and $H$ swapped. Thus $\Phi$ is bijective.

Third, prove $\Phi$ is a homomorphism (your check is right).

  • $\begingroup$ sorry, I'm still new at proofs. What does it mean for a map to be well defined? $\endgroup$ – yoshi Nov 19 '17 at 15:21
  • 1
    $\begingroup$ @yoshi You have to verify that $\phi\alpha\phi^{-1}$ is an automorphism of $H$ to begin with. Easy, but necessary. $\endgroup$ – egreg Nov 19 '17 at 16:00

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