I've heard this fact before but wasn't sure whether it pertained to an individual inner automorphism or $Inn(G)$, the set of all inner automorphisms of G.
I'm aware that if $g \in Z(G)$, then $\phi_g$ is the identity function ($\phi_g(x) = gxg^{-1} = xgg^{-1} = x$), but wasn't sure whether I could use this fact to conclude that the kernel is therefore the center of the group. Would appreciate help as to whether I'm on the right or wrong track here.
Best Answer
The center is almost by definition elements to which the associated inner automorphism is trivial (as you mention.)
What this means is that $G/Z(G) \cong \mathrm{Inn}(G)$. This can be seen by taking the homomorphism $G \to \mathrm{Inn}(G):g \mapsto \phi_g$, where $\phi_g(x)= gxg^{-1}$, and noting that its kernel is exactly the center. You can then apply the first isomorphism theorem.
There are two concepts going on simultaneously. Any automorphism $\phi:G \to G$ has trivial kernel by assumption. However, the set of maps $\{ \phi: G \to G\}$ can be given a group structure by composition: $\phi \circ\rho$ is associative, there is an identity, and inverses since all the maps are assumed to be automorphisms. We denote this group $\mathrm{Aut}(G)$. One subgroup we care about in particular is $\mathrm{Inn}(G)$ -- These are automorphisms of the form $\phi_g:G \to G$ given by $\phi_g(x)=g x g^{-1}$ for some $g \in G$.
Now, there is another homomorphism $\Phi:G \to \mathrm{Aut}(G)$ given by $g \mapsto \phi_g$. The image of this homomorphism is precisely the subgroup $\mathrm{Inn}(g)$
$\mathrm{ker}(\Phi)=Z(G)$ since it consists of group elements, so that conjugating by them is the identity map (trivial automorphism in the group.)
In other words, $g \in \mathrm{ker}(\Phi) \iff \phi_g=id$, or in other words, $gxg^{-1}=x$ for all $x \in G$.