Abstract algebra: Proof that if $\mathit{H}$ is a subgroup of index 2 in a finite group G, then gH=Hg for all g $\in$ G

Proof that if $\mathit{H}$ is a subgroup of index 2 in a finite group G, then g$\mathit{H}$=$\mathit{H}$g for all g $\in$ $\mathit{G}$.

I do know that index, or $\frac{|G|}{|H|}=2$ implies that $\mathit{H}$ is practically half of $\mathit{G}$, loosely speaking.

Moreover, I do know that every cyclic group of prime order is abelian, or gH=Hg.

Lastly, I do know that to show gH=Hg, it is suffice to show that $gH \subseteq Hg$ and $Hg \subseteq gH$.

But I could not tie those clues together and come up with an argument.

Could you please point me in the right direction?

Thanks.

Let $xH$ be a coset of $H$ in $G$. Since cosets partition $G$, either $xH=H$ or it is the other coset $G-H$ (the other coset is made of the leftovers, so it's the set complement of $H$). If $xH=H$, then $x\in H$ so $xH=H=Hx$. Otherwise $x\not\in H$, so $Hx \not= H$. Thus $xH=G-H=Hx$. So all the left and right cosets match.

• Thank you very much. It is all clear now. – user101998 Mar 17 '14 at 21:35

$G = H \cup gH = H \cup Hg$. Since cosets are disjoint and these are finite sets, it follows that $gH = Hg$.

• You don't even need that the sets are finite (complements are unique in all sets). – Tobias Kildetoft Mar 17 '14 at 8:15
• Thanks Tobias. You are right -- the proof works for any group with a subgroup of index 2. – user44441 Mar 18 '14 at 1:00

We know that the cosets partition the group. That is we have that:

$G=H\cup gH$ for some $g\not\in H$ and $H\cap gH=\emptyset$

Now we want to show that $\forall k\in G$ we have $k^{-1}Hk=H$

So we know that either $k^{-1}Hk=H$ or $k^{-1}Hk=gH$

But as $H$ is a subgroup we know that $1\in H$ and so $k^{-1}1k=1\in k^{-1}Hk$.

This shows that $k^{-1}Hk\cap H\neq \emptyset$ and and so $k^{-1}Hk=H$ as required

• Thank you for your answer. – user101998 Mar 17 '14 at 21:34