Is the kernel of an inner automorphism the center? 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.
 A: 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 $x \in G$. 
A: Given $g\in G$, conjugation by $g$ induces an automorphism $c_g$ of $G$ which maps an arbitrary element $x$ to $gxg^{-1}$. 
We then have a homomorphism $\varphi$ from $G$ into $Aut(G)$ which sends $g$ to $c_g$. The image of $G$ is the group of inner automorphisms.
So, when we speak of an inner automorphism of $G$, we mean a map $c_g$ for some $g\in G$. The kernel of any such map is trivial, being an automorphism.
However, the kernel of $\varphi$ is $Z(G)$. That is, the group of inner automorphisms is isomorphic to $G/Z(G)$.
