Subgroup of Index 2 is Normal

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Theorem

A subgroup of index 2 is always normal.


Proof

Suppose $H \le G$ such that $\left[{G : H}\right] = 2$.

Thus $H$ has two left cosets (and two right cosets) in $G$.

If $g \in H$, then $g H = H = H g$.

If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.

For the same reason, $g \notin H \implies H g = G \setminus H$.

That is, $g H = H g$.

The result follows from the definition of normal subgroup.

$\blacksquare$


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