# If Squares of Elements in a Group Lie in a Subgroup, then It is a Normal Subgroup

## Problem 469

Let $H$ be a subgroup of a group $G$.
Suppose that for each element $x\in G$, we have $x^2\in H$.

Then prove that $H$ is a normal subgroup of $G$.

(Purdue University, Abstract Algebra Qualifying Exam)

## Proof.

To show that $H$ is a normal subgroup of $G$, we prove that
$ghg^{-1}\in H$ for any $g\in G$ and $h\in H$.

For any $g\in G$ and $h\in H$ we have
\begin{align*}
&ghg^{-1}\\
&=g^2g^{-1}hg^{-1} &&\text{since $g=g^2g^{-1}$}\\
&=g^2g^{-1}hg^{-1}hh^{-1} &&\text{since $e=hh^{-1}$}\\
&=g^2(g^{-1}h)^2h^{-1}. \tag{*}
\end{align*}

It follows from the assumption that the elements $g^2$ and $(g^{-1}h)^2$ are in $H$.
Since $h\in H$, the inverse $h^{-1}$ is also in $H$.
Thus the expression in (*) is the product of elements in $H$, hence it is in $H$.

Thus, we have proved that $ghg^{-1}\in H$ for all $g\in G$, $h\in H$.
Therefore, the subgroup $H$ is a normal subgroup in $G$.

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