Subgroup Containing all Squares of Group Elements is Normal

Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$ with the property that:

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

Then $H$ is normal in $G$.

Corollary

The quotient group $G / H$ is abelian.

Proof

We have:

\(\ds \paren {x h}^2 h^{-1} \paren {x^{-1} }^2\)

\(=\)

\(\ds x h x h h^{-1} x^{-1} x^{-1}\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds x h x x^{-1} x^{-1}\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds x h x^{-1}\)

Group Axiom $\text G 3$: Existence of Inverse Element

Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both elements of $H$.

Thus:

$x h x^{-1} \in H$

and so $H$ is normal in $G$ by definition.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $13$