# 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

• Fundamental Theorem of Finitely Generated Abelian Groups and its application In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem. Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]
• 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 […]
• 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...

##### 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