# Any Subgroup of Index 2 in a Finite Group is Normal ## Problem 16

Show that any subgroup of index $2$ in a group is a normal subgroup. Add to solve later

Contents

## Hint.

1. Left (right) cosets partition the group into disjoint sets.
2. Consider both left and right cosets.

## Proof.

Let $H$ be a subgroup of index $2$ in a group $G$.
Let $e \in G$ be the identity element of $G$.

To prove that $H$ is a normal subgroup, we want to show that for any $g\in G$, $gH=Hg$.
If $g \in H$, then this is true. So we assume that $g \not \in H$.

Note that left cosets partition $G$ into two disjoint sets since the index is $2$.
Since $g \not \in H$, these are $eH$ and $gH$. (If $gH=H$, then $g \in H$.)

Similarly right cosets partition $G$ into two disjoint sets.
These disjoint right cosets are $He$ and $Hg$.

Because of these partitions, we have as sets
$gH=G – eH=G-H=G-He=Hg.$ Therefore $H$ is a normal subgroup in $G$. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Group Theory ##### The Center of a p-Group is Not Trivial

Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$. (Such a group is called a $p$-group.)...

Close