Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $Aut(H)$

share|improve this question

3 Answers 3

up vote 1 down vote accepted

You need to show that given an isomorphism $f:G \to H$, for each automorphism on $G$ you can find a unique isomorphism on $H$ (thus giving a $1-1$ mapping). For this just use the isomorphism to take the automorphism on $G$ and define an automorphism on $H$ by mapping elements (and confirm this actually gives an automorphism on $H$). Then uniqueness of the elements mapped will follow automatically, and you have a $1-!$ mapping in the other direction given by considering the inverse map $f^{-1}$. Then you can either prove that the two mappings between automorphisms are inverse to each other, or just appeal to the theorem that two sets are in $1-1$ correspondance if you can find an injective mapping from each set to the other.

share|improve this answer

Let $\theta : G \to H$ be an isomorphism. Define a map $\Phi : \operatorname{Aut}(G) \to \operatorname{Aut}(H)$ by $\Phi(f) = \theta \circ f \circ \theta^{-1}$. (Show that $\Phi(f)$ is indeed an automorphism of $H$.)

The map $\Phi$ is easily seen to be a homomorphism. It has the inverse $\Psi(g) = \theta^{-1} \circ g \circ \theta$. It follows that $\Phi$ is an isomorphism.

share|improve this answer

What you are trying to prove is true, but you can't focus on just two elements of each group, you need to look at the whole groups.

A composition is a good idea, but $\theta\circ \phi$ goes from $G$ to $H$, whereas you're looking for a map from $H$ to $H$. What can you do with a map from $G$ to turn it into a map from $H$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.