# New subgroup has twice the size of the old one?

While going through my class notes, I came across a statement I copied down from the board that I don't quite understand. The statement is this:

Let $H$ be a subgroup of a group $G$. Suppose that $g \in G$ but $g \notin H$, and that $g$ has order $2$ in $G$, and moreover $gHg^{-1} = H$. Then if $|H| = n$, the subgroup of $G$ generated by $H$ and $g$ has size $2n$.

I'm trying to do this on my own by showing that the only cosets of $H$ in $\langle g,H \rangle$ are $H$ and $gH$, and I guess the condition that $g$ has order $2$ tells us we don't get any extra cosets of the form $g^nH$, but I'm not seeing how to use the condition that $g$ conjugates $H$ into itself to show that there are no other cosets. Any help is appreciated!

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Note that you can always write $\,hg=g(ghg)=gh'\,\,,\,\,h,h'\in H\,$ , by the condition $\,H= gHg^{-1}=gHg\,$ , so that any word in $\,\langle g,H\rangle\,$ is of the form $\,g^ih\,\,,\,,i\in\{0,1\}\,\,,\,h\in H\,$ –  DonAntonio Dec 27 '12 at 4:00

HINT: Show that $H\cup gH$ is a subgroup of $G$.

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@AlexanderGruber: Thanks for catching the typo. –  Brian M. Scott Dec 27 '12 at 4:04
I was able to do it this way, thanks! Does this method generalize to the case where $g$ has arbitrary order? –  Jonas Dec 27 '12 at 4:11
@Jonas: You’re welcome. I’ve not really thought about it, but my immediate reaction is no: this seems to depend on using $g^{-1}=g$ to get $gH=Hg$. –  Brian M. Scott Dec 27 '12 at 4:15
@Jonas: Suppose that $g$ has order $n$. You can show that when $gHg^{-1} = H$, then $H \cup gH \cup g^2H \cup \ldots \cup g^{n-1}H$ is indeed a subgroup of $G$, but this does not have order $n|H|$ unless you assume $\langle g \rangle \cap H = \{1\}$. See my answer –  Mikko Korhonen Dec 28 '12 at 15:20

Assume that $gHg^{-1} = H$ and that $g \not\in H$ has order $2$. Now $\langle g \rangle \leq N_G(H)$, so $\langle g \rangle H \leq N_G(H)$ because $H \trianglelefteq N_G(H)$. Hence $\langle g, H \rangle = \langle g \rangle H$ and by the product formula this subgroup has order $2|H|$ because $\langle g \rangle \cap H = \{1\}$.

More generally, if $gHg^{-1} = H$, then $\langle g, H \rangle = \langle g \rangle H$ is a subgroup of order $$\frac{|g||H|}{|\langle g \rangle \cap H|}.$$

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