How can the stabilizer of a group element have more than one element itself? I suspect I've misunderstood the concept of a stabilizer.
Take a group $G$ and some $g \in G$. Now consider $s_1, s_2 \in Stab(g)$. The stabilizer is defined such that $s_1 g = s_2 g = g$, but if we operate on the right by $g^{-1}$, we obtain $s_1 = s_2 = e$ where $e$ is the identity. Thus $Stab(g) = \{e\}$.
This is clearly not correct, so I suspect I've failed to fully understand the concept. Where did I go wrong?
 A: The problem is that $g^{-1}g$ is not necessarily $e$! The action of $G$ on itself may be very different from the group law (think about conjugation, for example : in this case, $g^{-1}.g=g^{-1}gg=g\neq e$ in general). 
A: The stabilizer is associated with a group action. Actions aren't guaranteed to have inverses, so acting on the right by $g^{-1}$ is nonsensical. 
In general, given some group $G$ acting on a set $X$, the stabilizer is the elements of $G$ that act as the identity on $X$. Consider the action given by $h\mapsto ghg^{-1}$ (conjugation by $g$). This will be invariant under the action if $ghg^{-1}=h$, or if $gh=hg$, so if $g$ commutes with $h$. It follows that the center of $G$ stabilizes the above action. 
A: You can talk about stabilizer of an element of $G$ if the there is some group $H$ acting on $G$. If $G\times S\longrightarrow S$ is an action of $G$, the stabilizer of an element $s\in S$ is the subgroup $$G_{s}=\{g\in G:gs=s\}.$$
When $S$ is a group, then $gs=s$ implies $g=1_{G}$.
