a subgroup of index 2 is normal

Lemma.

Let $\mathrm{(}G\mathrm{,}\mathrm{\cdot }\mathrm{)}$ 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:

$$H,g_{1}H$$

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 $g_1{g}_{2}H\ne g_{1}H$ (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 $gh{g}^{-1}\in H$ and we are done. Otherwise, assume that $g\notin H$. Thus $gh\notin H$ and by the remark above $gh{g}^{-1}=(gh)g^{-1}\in H$, as desired. ∎

Title	a subgroup of index 2 is normal
Canonical name	ASubgroupOfIndex2IsNormal
Date of creation	2013-03-22 15:09:25
Last modified on	2013-03-22 15:09:25
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	5
Author	alozano (2414)
Entry type	Theorem
Classification	msc 20A05
Related topic	Coset
Related topic	QuotientGroup
Related topic	NormalityOfSubgroupsOfPrimeIndex