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Prove that $H$ is a normal subgroup of $G$ if and only if $\{g^{-1}h^{-1}gh|g \in G, h\in H\}$ is a subgroup of $H$.

Attempt: For the necessary condition, assume $H$ is normal in $G$. Then $g^{-1}h^{-1}g \in H$ for all $h \in H$. Then elements of the form $g^{-1}h^{-1}gh$ make a subgroup of $H$ by closure.

Beginner
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Chuck Franklin
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  • Can you please share your thoughts and efforts towards a proof? For example, what condition are you using for normality, and can you see any relations between it and elements of this particular form? –  Nov 13 '13 at 03:14
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    I think you mean $\langle g^{-1}h^{-1}gh|g\in G,h\in H \rangle$. – Alexander Gruber Nov 13 '13 at 03:18

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Let $H$ be a normal subgroup of $G$. Then $g^{-1}h^{-1}gh=(g^{-1}h^{-1}g)h$ is the product of two elements of $H$ and thus is contained in $H$. Therefore, the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is contained in $H$. We want to show that the subgroup generated by this set is a subgroup of $H$. To do this, it suffices to consider the product of two elements of the form $g^{-1}h^{-1}gh$, for $g\in G$ and $h\in H$: $$\begin{eqnarray*} g_1^{-1}h_1^{-1}g_1h_1 g_2^{-1}h_2^{-1}g_2h_2&=& (g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\\ \end{eqnarray*}$$ We have that $g_1^{-1}h_1^{-1}g_1$ and $g_2^{-1}h_2^{-1}g_2$ are elements of $H$ by normality, and of course $h_1$ and $h_2$ are elements of $H$, so $(g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\in H$. Therefore, $\langle g^{-1}h^{-1}gh|g\in G, h\in H\rangle$ is contained in $H$, so it is a subgroup of $H$.

Conversely, if the subgroup generated by the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is a subgroup of $H$, then $g^{-1}h^{-1}g\in H$ by closure for every $g\in G, h\in H$, so $H$ is normal in $G$.

Alexander Gruber
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  • Please note that though the problem says "$\ldots {g^{-1}h^{-1}gh|g\in G, h\in H}$ is a subgroup of $H\ldots$", this is false in the case that $H=G$. What it means is the subgroup generated by that set. – Alexander Gruber Nov 13 '13 at 03:35