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Stabilizer

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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).

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