$gH\subset Hg$ implies $gH=Hg$

Question

Let $G$ be a finite group and $H<G$. Let $g\in G$ such that $gH\subset Hg$. Prove that $gH=Hg$.

I know that for any group $G$ (finite of infinite), if for all $g\in G$, $g^{-1}Hg\subset H$ then $gH=Hg$. But here, $G$ is finite and $gHg^{-1}\subset H$ is true only for this particular $g$. Any hints?

$G$ is finite, so you can use cardinality as an argument here. What's the cardinality of $gH$ ? — Maxime Ramzi, Mar 13 '17 at 22:05
$|gH|=|H|=|Hg|$. I still don't get how to get the result from that :( — Cachorro, Mar 13 '17 at 22:08
if a set is contained in another set, and the cardinaility is equal what can you say ? — asddf, Mar 13 '17 at 22:12
What about this? $|gHg^{-1}|=|H|$ and $gHg^{-1}\subset H$ implies that $gHg^{-1}=H$ — Jax, Mar 13 '17 at 22:13

score 1 · Accepted Answer · answered Mar 13 '17 at 22:23

1

We know that $|gH|=|Hg|$. Since $G$ is finite $gH$ and $Hg$ are also finite. Now, $gH\subset Hg$ implies that $gH=Hg$.

answered Mar 13 '17 at 22:23

Jax

756

$gH\subset Hg$ implies $gH=Hg$

1 Answers1