Fun Exercise (Abstract Algebra): Let $G$ be a group, and let $H$ be a subgroup of $G$. Show that if $H$ is of index $2$ in $G$, i.e., $\left|G:H\right|=2$, then $H$ is normal.

Question

thomas5267 · Answer

I have no idea whether this is correct or not.

Since the index is 2, the group G can be partitioned into two sets, those in H and those not in H.  Let h in H in G.  Can we claim that hH=Hh?  I would say yes because we could use the bijection hH to H^(-1)h^(-1) and use the fact that there is a bijection between an element and its inverse.  Let g in G but not in H.  If gH != Hg then it must be the case that gH or Hg equals H since there are only two cosets.  Say if gH=H then g must be in H and this violates the assumption the g is in G but not in H?  Therefore gH=Hg and H is normal?

I have no idea what I am doing.

across · Answer

Awesome! If $h\in H$, then, since $H$ is a group, $hH=H=Hh$ immediately. The rest follows just like you wrote. :-) Here's your medal.

thomas5267 · Answer

OMG I got it!  I knew basically nothing about group theory and got it!