# Stabilizer

A stabilizer is a part of a monoid (or group) acting on a set.

Specifically, let $M$ be a monoid operating on a set $S$, and let $A$ be a subset of $S$. The stabilizer of $A$, sometimes denoted $\text{stab}(A)$, is the set of elements of $a$ of $M$ for which $a(A) \subseteq A$; the strict stabilizer' is the set of $a \in M$ for which $a(A)=A$. In other words, the stabilizer of $A$ is the transporter of $A$ to itself.

By abuse of language, for an element $x\in S$, the stabilizer of $\{x\}$ is called the stabilizer of $x$.

The stabilizer of any set $A$ is evidently a sub-monoid of $M$, as is the strict stabilizer. Also, if $a$ is an invertible element of $M$ and a member of the strict stabilizer of $A$, then $a^{-1}$ is also an element of the strict stabilizer of $a$, for the restriction of the function $a : S \to S$ to $A$ is a bijection from $A$ to itself.

It follows that if $M$ is a group $G$, then the strict stabilizer of $A$ is a subgroup of $G$, since every element of $G$ is a bijection on $S$, but the stabilizer need not be. For example, let $G=S= \mathbb{Z}$, with $g(s) = g+s$, and let $A=\mathbb{Z}_{>0}$. Then the stabilizer of $A$ is the set of nonnegative integers, which is evidently not a group. On the other hand, the strict stabilizer of $A$ is the set $\{0\}$, the trivial group. On the other hand, if $A$ is finite, then the strict stabilizer and the stabilizer are one and the same, since $a : S \to S$ is bijective, for all $a\in G$.

Proposition. Let $G$ be a group acting on a set $S$. Then for all $x\in S$ and all $a \in G$, $\text{stab}(ax) = a\, \text{stab}(x) a^{-1}$.

Proof. Note that for any $g \in \text{stab}(x)$, $(aga^{-1})ax = agx = ax.$ It follows that $$a\, \text{stab}(x) a^{-1} \subseteq \text{stab}(ax) .$$ By simultaneously replacing $x$ with $ax$ and $a$ with $a^{-1}$, we have $$\text{stab}(ax) \subseteq a\, \text{stab}(x) a^{-1} ,$$ whence the desired result. $\blacksquare$

In other words, the stabilizer of $ax$ is the image of the stabilizer of $x$ under the inner automorphism $\text{Int}(a)$.