Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
Neat and clear exposition! +1 –  amWhy Jan 29 '13 at 20:43
add comment

I think this is fine.

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.