# 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?

• Stabilizers are usually used in group actions, where a group acts on a set. In your example, that set is the group again, and the action is left multiplication. In this example, yes, each element has only $e$ as stabilizer. But group actions are much more general than this. See mathworld.wolfram.com/Stabilizer.html Aug 14 '16 at 0:55
• Usually, we consider $G$ to have a group action on a set $X$. The elements that we're finding stabilizers are just elements of $X$, so no notion of inverse applies. Aug 14 '16 at 0:57
• You misunderstand notations between group actions and left multiplication defined on G. Aug 14 '16 at 1:07

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).
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.
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 $g=1_{G}$.