# Cosets of center and Inner automorphisms

I'm supposed to prove that if $G$ is a group and $Z$ is the set of all $a \in G$ with $ax = xa$ for all $x \in G$ then the cosets of $Z$ in $G$ are isomorphic to the group of inner automorphisms of $G$ which is the set of all functions $x \mapsto gxg^{-1}$ for $g \in G$. So probably the isomorphism here is gonna be $gZ \mapsto (x \mapsto gxg^{-1})$ because what else would it be? Nothing. And so it's pretty clear that this here is a morphism and surjective but I get nowhere trying to prove it is injective. $axa^{-1} = bxb^{-1} \forall x\rightarrow ??? \rightarrow aZ = bZ$, in summary.

-
It's cleaner to use the first isomorphism theorem. You've written down a surjective homomorphism $G \to \text{Inn}(G)$. What is its kernel? – Qiaochu Yuan Dec 2 '12 at 10:25

Suppose that $axa^{-1}=bxb^{-1}$ for all $x\in G$; then $x=a^{-1}bxb^{-1}a=\left(b^{-1}a\right)^{-1}x\left(b^{-1}a\right)$ for all $x\in G$, so $\left(b^{-1}a\right)x=x\left(b^{-1}a\right)$ for all $x\in G$, and therefore $b^{-1}a\in Z$. There’s not much left; can you finish it from here?
Alternatively, you could back up and try a slightly less computational approach: for what $g\in G$ is it true that the map $x\mapsto gxg^{-1}$ is the identity? These $g$ would form the kernel of your mapping $\varphi$ from $G$ to $\operatorname{Inn}(G)$. Clearly $gxg^{-1}=g$ for all $x\in G$ iff $gx=xg$ for all $x\in G$ iff $g\in Z$. In other words, $\ker\varphi=Z$; what does that tell you about $G/Z$?
$b^{-1}a \in Z \rightarrow b^{-1}aZ = Z \rightarrow aZ = bZ$, yeah? – Toby Carter Dec 2 '12 at 10:31
@Tony: Yes. Alternatively, $b^{-1}a\in Z$ implies that $a\in bZ$, and since cosets are either identical or disjoint, that implies that $aZ=bZ$. – Brian M. Scott Dec 2 '12 at 10:32