Showing that any group has a conjugation automorphism How can I show that in any group, given two elements there is a conjugation automorphism that sends $g_1*g_2$ to $g_2*g_1$? I tried using the method for conjugation by a single element of the group, but that didn't give me the right answer...
 A: Write $\gamma_h$ for the map which sends $g\mapsto h^{-1}gh$ for all $g\in G$. So the map which corresponds to conjugation by the fixed element $h$. Then proving that this is an automorphism solves your problem, as taking $h=g_1$ we see that $\gamma_{g_1}(g_1g_2)=g_2g_1$.
I will now give you some hints so that you can prove yourself that is an automorphism.
Hint 1: Homomorphism. Use the fact that $h^{-1}k_1k_2h=h^{-1}k_1hh^{-1}k_2h$.
Hint 2: Injection. We have shown that $\gamma_h$ is a homomorphism, so if it is not an injection then there exists some non-trivial $g\in G$ such that $\gamma_h(g)=1$. Why can this not happen?
I think I'll leave you to prove that $\gamma_h$ is surjective. It isn't too hard. Note that if $G$ is finite then you have already done this (why?). However, there exist infinite groups $G$ with injective but non-surjective homomorphisms $G\rightarrow G$. For example, consider the integers under addition...
EDIT: In the comments below here and below the main question, Tony (the OP) has said that he was confused as he thought that the map $\gamma_h$ would map $g_1g_2$ to $g_2g_1$ for all $g_1, g_2\in G$. If such a map exists, it turns out that the group is abelian. To see this, begin by noting that for all $g\in G$ we have $\gamma_h(g^{-1}hg)=h$, but as $\gamma_h(g^{-1}hg)=h^{-1}g^{-1}hgh$ this means that $g^{-1}hg=h$, so $g$ and $h$ commute. As this holds for all $g\in G$, $h\in Z(G)$. Now, let $g_1$ and $g_2$ be arbitrary elements of $G$. Then $\gamma_h(g_1g_2)=g_2g_1$ by assumption. However, because $h$ is central $\gamma_h$ acts trivially, so $g_1g_2=g_2g_1$ as required.
