# Why does every isomorphic group has a conjugacy class for the given function? [closed]

I'm trying to teach myself with Sylow Theorems and conjugacy classes. I was reading a proof which involved automorphism and conjugacy class in the group and this was one of the step to the proof.

The statement I'm trying to "If $C$ is a conjugacy class of a group $G$ and $f$ is an automorphism of $G$, prove that $f(C)$ is also a conjugacy class of $G$" The proof consists of assuming $C$ is a conjugacy class of an element in $G$ and since $f$ is an automorphism which is also an isomorphism.

"every isomorphic group has a conjugacy class for the given function."

Why is this a true statement? What would be the steps to prove it?

## closed as unclear what you're asking by Derek Holt, user91500, zhoraster, Alex Provost, user223391 Apr 15 '17 at 15:11

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• I think there is more context needed here: What function, why isomorphic groups, what exactly are you trying to do? Remember that the exact wording of such a proof is not unique, every author might prove the result differently. – Dirk Apr 13 '17 at 6:22
• @Bemte I aplogize for being vague. Here is the statement of the problem "If $C$ is a conjugacy class of a group $G$ and $f$ is an automorphism of $G$, prove that $f(C)$ is also a conjugacy class of $G$" The proof consists of assuming $C$ is a conjugacy class of an element in $G$ and since $f$ is an automorphism which is also an isomorphism. – Ya G Apr 13 '17 at 6:44
• Please edit your post rather than putting corrections in comments. – Derek Holt Apr 13 '17 at 8:07

We must show that any two elements of $f(C)$ are conjugate, and that the conjugate of any element of $f(C)$ also belongs to $f(C)$.
Let $f(c)$ and $f(c')$ be generic elements of $f(C)$. Since $c$ and $c'$ are conjugate, $c = gc'g^{-1}$ for some $g \in G$, and so $f(c) = f(g)f(c')f(g)^{-1}$ as required.
Let $gf(c)g^{-1}$ be the conjugate of a generic element of $f(C)$. We may rewrite it as $gf(c)g^{-1} = f(f^{-1}(g)c(f^{-1}(g))^{-1})$, which shows that it belongs to $f(C)$.