# Normalizer of a subset

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

Let $N(S)= \{a \in G \mid aSa^{-1} = S \}$.

Prove $N(bSb^{-1})=bN(s)b^{-1}$ for a fixed $b\in G$.

Would this be the correct definition $N(bSb^{-1}) = \{ a \in G \mid abSb^{-1}a^{-1} = bSb^{-1} \}$ to start from?

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## 2 Answers

Let $g\in N(bSb^{-1})$ in which as you assumed, $b\in G$ is a fixed element. So according to the definition, $$g(bSb^{-1})=(bSb^{-1})g$$ iff $$(gb)Sb^{-1}=bS(b^{-1}g)$$ iff $$b^{-1}(gb)S=S(b^{-1}g)b$$ iff $$(b^{-1}gb)S=S(b^{-1}gb)$$ This means that $b^{-1}gb\in N(S)$ iff $g\in bN(S)b^{-1}$. Here you are.

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Neat and clear exposition! +1 –  amWhy Jan 29 '13 at 20:43

I think this is fine.

If $a S a^{-1} = S$ then $bab^{-1} \times bSb^{-1} \times bab^{-1} = b \times aSa^{-1} \times b^{-1} = bSb^{-1}$

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