TITLE: if $H$ is a unique subgroup of order $d$ then $H$ is a normal subgroup
QUESTION [5 upvotes]: Let $H$ be a subgroup of a group, $G$ such that $H$ is the only subgroup in $G$ with order $d$.
Prove that $H\vartriangleleft G$.
I have to questions:
1) as you see it is not given that $G$ is finite. is it true that $H\vartriangleleft G$ even if $G$ is not finite?
2) suppose that $G$ is finite. then my try was to take $h\in H$, $g\in G$.
now, $(ghg^{-1})^d=gh^dg^{-1}=gg^{-1}=e$ and I tried to show by contradiction that $d$ must be the order of $ghg^{-1}$: if $1\leq k<d$ and $(ghg^{-1})^k=e$ then $(ghg^{-1})^k=gh^kg^{-1}=e$ $\Longrightarrow h^k=e$ so the order of $h$ must divide $k$ which is smaller then $d$ so it seemed reasonable for me that the contradiction should arrise from the fact that the order of $h$ must be $d$ i.e. $H$ is cyclic.
I looked up similar questions and found that if $G$ is finite and for every divisor of $|G|$ there is a unique subgroup of the order of that divisor then $G$ is cyclic and in my case it would mean that $H$ is also cyclic and hence the contradiction. but I'm not given that $G$ is finite and that for every divisor of $|G|$ there is a unique subgroup of that order. so I'm kind of stuck here

REPLY [5 votes]: Note that for all $g\in G$, $gHg^{-1}$ is a subgroup of order $d$. Thus $gHg^{-1} = H$ for all $g\in G$ and so $H$ is normal.

REPLY [3 votes]: If $g\in G$ then $gHg^{-1}$ is a subgroup of same order as $H$, hence by assumption $gHg^{-1}=H$ for all $g\in G$.