Prove N(S) is a subgroup 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.

$eSe^{-1}=eSe=S$, 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.