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:
\(\ds \forall x \in G: \, \) | \(\ds \paren {x^2} H\) | \(=\) | \(\ds \paren {x H}^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds H\) | as $G / H$ is of order $2$ |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $18 \ \text {(i)}$