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Let $S$ be a non empty subset of a group $G$. if $a\in G$, let $aSa^{-1}=(asa^{-1}|s\in S)$.

Let $N(S)=[a \in g|aSa^{-1} = S]$.

Prove $N(S)$ is a subgroup of $G$.

I'm really having difficulty grasping what this subspace consists of, I understand so far $N(S)=[a \in g|asa^{-1} = S|s \in S]$ but am having trouble grasping the form of an element in this subgroup.

Any help would be appreciated.

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This is also called the normalizer of $S$, i.e., it is the largest subgroup of $G$ such that $S$ is normal in $N(S)$. Hopefully that helps clears up the definition. –  Clayton Jan 29 '13 at 3:44
    
that notation sure is confusing –  bobdylan Jan 29 '13 at 3:55
    
It's helpful because there is another subgroup with a similar nature: the centralizer of $S$ (the largest subgroup of $G$ such that all elements in $C(S)$ commute with all elements in $S$). Both of these come up naturally when talking about operations of a group on a set. –  Clayton Jan 29 '13 at 4:03
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2 Answers

up vote 3 down vote accepted

$eSe^{-1}=eSe=S$, so $e\in N(S)$ and $N(S)$ is nonempty.

Let $x,y\in N(S)$. Then $xSx^{-1}=ySy^{-1}=S$, so

$(xy^{-1})S(xy^{-1})^{-1}$

$=(xy^{-1})(ySy^{-1})(yx^{-1})$

$=x(y^{-1}y)S(y^{-1}y)x^{-1}$

$=xSx^{-1}$

$=S$

so that $xy^{-1}\in N(S)$.

This shows that $N(S)$ is a subgroup.

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This proof implicitly assumed that $y\in N(S)\Rightarrow y^{-1}\in N(S)$. I have added a step in the line of equalities to fix this. –  Aaron Jan 29 '13 at 4:18
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Here is a higher-level proof just in case.

Let $Aut(G)$ be the automorphism group of $G$. The set $\Sigma$ of all automorphisms that leave $S$ invariant is a subgroup of $Aut(G)$. Let $Inn: G \to Aut(G)$ given by $Inn(g)(x)=g x g^{-1}$. Then $Inn$ is a homomorpishm. In that context, $N(S)$ is the just the inverse image under $Inn$ of $\Sigma$ and so is a subgroup.

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