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A '''stabilizer''' is a part of a [[monoid]] (or [[group]]) [[group action|acting]] on a set.

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

By abuse of language, for an element <math>x\in S</math>, the stabilizer of <math>\{x\}</math> is called the stabilizer of <math>x</math>.

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

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

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

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

In other words, the stabilizer of <math>ax</math> is the image of the stabilizer of <math>x</math> under the [[inner automorphism]] <math>\text{Int}(a)</math>.

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== See also ==

* [[Orbit-stabilizer theorem]]

[[Category:Abstract algebra]]
[[Category:Group theory]]