# a subgroup of index 2 is normal

###### Lemma.

Let $(G,\cdot)$ be a group and let $H$ be a subgroup   of $G$ of index 2. Then $H$ is normal in $G$.

###### Proof.

Let $G$ be a group and let $H$ be an index 2 subgroup of $G$. By definition of index, there are only two left cosets  of $H$ in $G$, namely:

 $H,\quad g_{1}H$

where $g_{1}$ is any element of $G$ which is not in $H$. Notice that if $g_{1},\ g_{2}$ are two elements in $G$ which are not in $H$ then $g_{1}\cdot g_{2}$ belongs to $H$. Indeed, the coset $g_{1}g_{2}H\neq g_{1}H$ (because $g_{1}g_{2}=g_{1}h$ would immediately yield $g_{2}=h\in H$) and so $g_{1}g_{2}H=H$ and $g_{1}g_{2}\in H$.

Let $h\in H$ be an arbitrary element of $H$ and let $g\in G$. If $g\in H$ then $ghg^{-1}\in H$ and we are done. Otherwise, assume that $g\notin H$. Thus $gh\notin H$ and by the remark above $ghg^{-1}=(gh)g^{-1}\in H$, as desired. ∎

Title a subgroup of index 2 is normal ASubgroupOfIndex2IsNormal 2013-03-22 15:09:25 2013-03-22 15:09:25 alozano (2414) alozano (2414) 5 alozano (2414) Theorem msc 20A05 Coset QuotientGroup NormalityOfSubgroupsOfPrimeIndex