Difference between revisions of "Stabilizer"

Latest revision as of 17:47, 9 September 2008

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 $\text{Int}(a)$ .

This article is a stub. Help us out by expanding it.

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-A '''stabilizer''' is a part of a [[monoid]] (or [[group]]) acting on a set.
+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 integers, 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>.
+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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Difference between revisions of "Stabilizer"

Latest revision as of 17:47, 9 September 2008

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