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Let $G$ be a group and $H$ subgroup of $G$, $N(H):=\{g\in G; gHg^{-1}=H\}$

I need to prove that $N(H)$ is subgroup of $G$.

It's almost the same question like :How to show $\forall g \in G, gHg^{-1} = H \Leftrightarrow \forall g \in G, gHg^{-1} \subseteq H$?

But I need the inverse direction of the answer.

Attempt: The inverse of $gHg^{-1}$ is:

$$\begin{align}(gHg^{-1})^{-1}&=(g^{-1})^{-1}Hg^{-1}=gHg^{-1}=H \\&\implies (gHg^{-1})^{-1}=gHg^{-1} \\&\implies \text{the inverse of element is the element}\end{align}$$

Is it true?

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  • $\begingroup$ The inverse of a set!? Be careful, what you need to prove is that the inverse of each element that lives in $N(H)$ lives in $N(H)$ too, $\endgroup$
    – Daniel
    May 25, 2015 at 14:43

2 Answers 2

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The elements of $N(H)$ aren't $gHg^{-1}$ but $g \in G$ such that $H$ has that property.

If $g, g' \in N(H)$ then $$gg' H (gg')^{-1} = g(g' H g'^{-1})g^{-1} = gHg^{-1} = H$$ then $g\dot \,g' \in N(H)$

If $g \in N(H)$ then $g^{-1} H (g^{-1})^{-1} = g^{-1} H g = H$, because $gHg^{-1} = H$, then $g^{-1} \in N(H)$.

And you're done.

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You will need that $H$ is subgroup of $G$. Note that if $e\in G$ is the identity, then $e\in N(H)$ because $eHe = H$. Next, you want to show that if $g\in N(H)$ then $g^{-1} \in N(H)$. Equivalently, for all $h\in H$, $(ghg^{-1})^{-1}\in H$. But $(ghg^{-1})^{-1} = gh^{-1}g^{-1}$ is indeed in $H$ because $h^{-1} \in H$ and $g\in N(G)$.

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