Automorphisms are isomorphisms

[Dustinsfl]

If \sigma\in Aut(G) and \varphi_g is conjugation by g prove \sigma\varphi_g\sigma^{-1}=\varphi_{\sigma(g)}. Deduce Inn(G)\trianglelefteq Aut(G)

Let x\in G.

\sigma\varphi_g\sigma^{-1}(x)=\sigma(g\sigma^{-1}(x)g^{-1})=\sigma(g)x\sigma(g)^{-1}

Why is this:
\sigma(g\sigma^{-1}(x)g)=\sigma(g)x\sigma(g)^{-1}

 

[Fredrik]

If \sigma\in Aut(G) and \varphi_g is conjugation by g prove \sigma\varphi_g\sigma^{-1}=\varphi_{\sigma(g)}. Deduce Inn(G)\trianglelefteq Aut(G)

You should include more information when you ask questions like this. Perhaps I suck more than most at abstract algebra terminology, but I have no idea what Inn and trianglelefteq means, and I'm not sure "conjugation" means what I'm guessing it means either.

Let x\in G.

\sigma\varphi_g\sigma^{-1}(x)=\sigma(g\sigma^{-1}(x)g)=\sigma(g)x\sigma(g)^{-1}

OK, it looks like my guess about "conjugation" was right. You meant \varphi_g(h)=ghg^{-1}, right? (You missed a ^{-1}).

Why is this:
\sigma(g\sigma^{-1}(x)g)=\sigma(g)x\sigma(g)^{-1}

Because Aut(G) is the group of automorphisms on G. Automorphisms are permutations that preserve the group multiplication operation.

 

[Dustinsfl]

You should include more information when you ask questions like this. Perhaps I suck more than most at abstract algebra terminology, but I have no idea what Inn and trianglelefteq means, and I'm not sure "conjugation" means what I'm guessing it means either. OK, it looks like my guess about "conjugation" was right. You meant \varphi_g(h)=ghg^{-1}, right? (You missed a ^{-1}).Because Aut(G) is the group of automorphisms on G. Automorphisms are permutations that preserve the group multiplication operation.

Inn is inner automorphism. Correct on conjugation.
trianglelefteq is a normal subgroup.

I still don't see how that is equal.

 

[Fredrik]

I assume that you know that automorphisms are isomorphisms, that isomorphisms are homomorphisms, and that every homomorphism f satisfies f(xy)=f(x)f(y) for all x,y in the group? So what is f(xyz)?

What you actually need to evaluate is of the form f(xyz^-1), but it's easy to prove that f(x^-1)=f(x)^-1 for all x in the group.