If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order
Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup.
Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.
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Proof.
The index of the subgroup $H$ in $G$ is $2$, hence $H$ is a normal subgroup.
(See the post “Any Subgroup of Index 2 in a Finite Group is Normal“.)
Also, the order of $H$ must be odd, otherwise $H$ contains an element of order $2$.
So it remains to prove that $H$ is abelian.
Let $a\in S$ be an element of order $2$.
As $a\notin H$, the left coset $aH$ is different from $H$.
Since the index of $H$ is $2$, we have $aH=G\setminus H=S$.
So for any $h\in H$, the order of $ah$ is $2$.
It follows that we have for any $h\in H$
\[e=(ah)^2=ahah,\]
where $e$ is the identity element in $G$.
Equivalently, we have
\[aha^{-1}=h^{-1} \tag{*}\]
for any $h\in H$.
(Remark that $a=a^{-1}$ as the order of $a$ is $2$.)
Using this relation, for any $h, k \in H$, we have
\begin{align*}
(hk)^{-1}&\stackrel{(*)}{=} a(hk)a^{-1}\\
&=(aha^{-1})(aka^{-1})\\
&\stackrel{(*)}{=}h^{-1}k^{-1}=(kh)^{-1}.
\end{align*}
As a result, we obtain $hk=kh$ for any $h, k$.
Hence the subgroup $H$ is abelian.
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