Subgroup of Index 2 contains all Squares of Group Elements

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Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$ whose index is $2$.


Then:

$\forall x \in G: x^2 \in H$


Proof

By Subgroup of Index 2 is Normal, $H$ is normal in $G$.

Hence the quotient group $G / H$ exists.

Then we have:

\(\, \displaystyle \forall x \in G: \, \) \(\displaystyle \paren {x^2} H\) \(=\) \(\displaystyle \paren {x H}^2\)
\(\displaystyle \) \(=\) \(\displaystyle H\) as $G / H$ is of order $2$

$\blacksquare$


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