TITLE: What does the isomorphism $G / Z(G) \cong \text{Inn}(G)$ mean?
QUESTION [9 upvotes]: $G$ is a group. $Z(G) = \{g \in G: \forall x \in G (gx = xg) \}$ is called the center of $G$. It is easy to verify that $Z(G) \triangleleft G$. $\text{Inn}(G)$ is the inner automorphism group of $G$. As a consequence of the first isomorphism theorem, we have
$$G / Z(G) \cong \text{Inn}(G).$$ 
The homomorphism involved here is defined as $a \in G \mapsto \sigma_{a} \in \text{Inn}(G)$ where $\sigma_{a}$ is a bijection from $G$ to $G$ with $\sigma_{a}(x) = axa^{-1}$. The details can be found here: Factor Group over Center Isomorphic to Inner Automorphism Group.
The isomorphism is too abstract. Could anyone offer some intuitive (or even visual) explanations?

REPLY [3 votes]: I would suggest looking at it in the following way.
If we define $Aut(G)$ to be the group of automorphisms of $G$, i.e.
$$
Aut(G) = \{f : G \to G \mid f \text{ is a bijective homomorphism}\}
$$
then there is a map $\phi : G \to Aut(G)$ given by
$$
\phi(g) = \big(h \mapsto ghg^{-1}\big)
$$
that is, $\phi(g)$ is the automorphism given by conjugation by $g$.
Now, it is (hopefully) clear that the image of $G$ in $Aut(G)$ is exactly the set of inner automorphisms, i.e. $Inn(G)$. What is the kernel? Well, the kernel is exactly those $g \in G$ such that $h = ghg^{-1}$ for all $h \in G$. i.e. the kernel is $Z(G)$!
Thus it follows that $im(\phi) = G / \ker\phi$ (this is a restatement of the first isomorphism theorem), i.e. $Inn(G) = G / Z(G)$.