Say $G \cong H$ are isomorphic groups. Show $Aut(G) \cong Aut(H)$
I just made this up so I'm not sure if actually $Aut(G) \cong Aut(H)$ is true but I'm $99.9\%$ sure this should be true
I'm having trouble with the proof :'(
Let $\theta:G \rightarrow H$ be the isomorphism and $\phi\in Aut(G)$
Let $g_1,g_2\in G$ and let $h_1,h_2 \in H \ \ with \ \ h_1 = \theta(g_1) \ \ and \ \ h_2 = \theta(g_2)$
If $\phi(g_1) = g_2$ I want to show that $\exists \alpha\in Aut(H)$ such that
$\alpha(h_1) = h_2$
I tried composing $\theta$ and $\phi$ to get into $H$ but that doesn't get me anywhere. Is there something wrong with this approach?
Thanks ! :D
Wait a minute....
$\theta(\phi(g_1)) = \theta(g_2) = h_2$
$\theta(\phi(g_2)) = \theta(g_1) = h_1$
I think I see something here, $h_1$ and $h_2$ were permuted in the same way $g_1$ and $g_2$ were but $\theta \phi : G\rightarrow H$ so it cant be in $Aut(H)$