Check center of group proof

Question-Let G be a group and let g∈G. If z ∈ Z(G) (the center of a group G), show that the inner automorphism induced by g is the same as the inner automorphism induced by zg(that is ϕg = ϕzg) Hint: What is the definition of Z(G)?

Proof:

Since Z(G) just contains e, then z must equal e

so ϕg = ϕzg

ϕg=ϕeg

ϕg=ϕg

I'm sure I'm missing some stuff, but is this even close?

Re: Check center of group proof

Hint: z commutes with all elements of G.

Re: Check center of group proof

no, Z(G) can be lots bigger than just {e}. in fact, if G is abelian Z(G) is all of G!

however, for any z in Z(G), we have zx = xz for any element x of G. and this means that:

$\displaystyle \phi_z(x) = zxz^{-1} = xzz^{-1} = xe = x$ for any x in G, that is, $\displaystyle \phi_z$ is the identity automorphism of G.

you'll be done if you can prove that $\displaystyle \phi_g \circ \phi_h = \phi_{gh}$ for any g,h in G (why?)