Given $G$ is a finite group, $a$ and $b \in G$, $H\le G$, and $N=N(H)$, prove that if $aHa^{-1}=bHb^{-1}$, then $aN=bN$.

Question

I am having difficulties proving the below statement:

Given $G$ is a finite group, $a$ and $b \in G$, $H$ is a subgroup of $G$, and $N=N(H)$, i.e. $N$ is the normalizer of $H$:

If $aHa^{-1}=bHb^{-1}$, then $aN=bN$

I have generated four other pieces of information that I think may be useful:

1. $N(H) < G$
2. $H \subset N(H)$
3. $H \lhd N(H)$
4. $b^{-1}a \in N(H)$

I would like to structure this proof in to two sections; one that shows $aN \subseteq bN$ and one that shows $bN \subseteq aN$ (presumably without any loss of generality from the former section).

I have a hunch that an important part of this proof is recognizing that $n_1\circ H \circ n_1^{-1}$, where $n_1$ and $n_1^{-1} \in N$, is the same set as $H$...but I cannot quite figure out how to set it up.

Cheers.

Answer 1

The following statements are equivalent:

$$aHa^{-1}=bHb^{-1}$$ $$b^{-1}aHa^{-1}b=H$$ $$b^{-1}aH(b^{-1}a)^{-1}=H$$ $$b^{-1}a\in N$$ $$aN=bN$$