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. $\blacksquare$