# 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.

## 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.

### More from my site

• Any Subgroup of Index 2 in a Finite Group is Normal Show that any subgroup of index $2$ in a group is a normal subgroup. Hint. Left (right) cosets partition the group into disjoint sets. Consider both left and right cosets. Proof. Let $H$ be a subgroup of index $2$ in a group $G$. Let $e \in G$ be the identity […]
• Group of Order 18 is Solvable Let $G$ be a finite group of order $18$. Show that the group $G$ is solvable.   Definition Recall that a group $G$ is said to be solvable if $G$ has a subnormal series $\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G$ such […]
• Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable Let $p, q$ be prime numbers such that $p>q$. If a group $G$ has order $pq$, then show the followings. (a) The group $G$ has a normal Sylow $p$-subgroup. (b) The group $G$ is solvable.   Definition/Hint For (a), apply Sylow's theorem. To review Sylow's theorem, […]
• Abelian Group and Direct Product of Its Subgroups Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers. Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.   Hint. Consider […]
• The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ Let $p$ be a prime number. Let $G$ be a non-abelian $p$-group. Show that the index of the center of $G$ is divisible by $p^2$. Proof. Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$. Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
• Normal Subgroup Whose Order is Relatively Prime to Its Index Let $G$ be a finite group and let $N$ be a normal subgroup of $G$. Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$. (a) Prove that $N=\{a\in G \mid a^n=e\}$. (b) Prove that $N=\{b^m \mid b\in G\}$.   Proof. Note that as $n$ and […]
• Non-Abelian Group of Order $pq$ and its Sylow Subgroups Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$. Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.   Hint. Use Sylow's theorem. To review Sylow's theorem, check […]
• Are Groups of Order 100, 200 Simple? Determine whether a group $G$ of the following order is simple or not. (a) $|G|=100$. (b) $|G|=200$.   Hint. Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$. Check out the post Sylow’s Theorem (summary) for a review of Sylow's […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8

Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.

Close