If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd

Group Theory Problems and Solutions in Mathematics

Problem 326

Prove that if $G$ is a finite group of even order, then the number of elements of $G$ of order $2$ is odd.

 
FavoriteLoadingAdd to solve later

Sponsored Links

Proof.

First observe that for $g\in G$,
\[g^2=e \iff g=g^{-1},\] where $e$ is the identity element of $G$.
Thus, the identity element $e$ and the elements of order $2$ are the only elements of $G$ that are equal to their own inverse elements.

Hence, each element $x$ of order greater than $2$ comes in pairs $\{x, x^{-1}\}$.
So we have
\begin{align*}
&G=\\
&\{e\}\cup \{\text{ elements of order $2$ } \} \cup \{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\},
\end{align*}
where $x_i$ are elements of order greater than $2$ for $i=1,2, \dots, k$.

As we noted above, the elements $x_i, x_i^{-1}$ are distinct.
Thus the third set contains an even number of elements.

Therefore we have
\begin{align*}
&\underbrace{G}_{\text{even}}=\\
&\underbrace{\{e\}}_{\text{odd}}\cup \{\text{ elements of order $2$ } \}\cup \underbrace{\{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\}}_\text{even}
\end{align*}
It follows that the number of elements of $G$ of order $2$ must be odd.

If the Order of a Group is Even, then it has a Non-Identity Element of Order 2

The consequence of the problem yields that the number of elements of order $2$ is odd, in particular, it is not zero.

Hence we obtain:

If the order of a group is even, then it has a non-identity element of order 2.

FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

  • Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian GroupTorsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order. (a) Prove that $T(A)$ is a subgroup of $A$. (The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]
  • The Order of $ab$ and $ba$ in a Group are the SameThe Order of $ab$ and $ba$ in a Group are the Same Let $G$ be a finite group. Let $a, b$ be elements of $G$. Prove that the order of $ab$ is equal to the order of $ba$. (Of course do not assume that $G$ is an abelian group.)   Proof. Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is, \[(ab)^n=e, […]
  • Use Lagrange’s Theorem to Prove Fermat’s Little TheoremUse Lagrange’s Theorem to Prove Fermat’s Little Theorem Use Lagrange's Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat's Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.   Before the proof, let us recall Lagrange's Theorem. Lagrange's Theorem If $G$ is a […]
  • Fundamental Theorem of Finitely Generated Abelian Groups and its applicationFundamental 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 […]
  • Group of Order $pq$ Has a Normal Sylow Subgroup and SolvableGroup 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, […]
  • Group Homomorphism Sends the Inverse Element to the Inverse ElementGroup Homomorphism Sends the Inverse Element to the Inverse Element Let $G, G'$ be groups. Let $\phi:G\to G'$ be a group homomorphism. Then prove that for any element $g\in G$, we have \[\phi(g^{-1})=\phi(g)^{-1}.\]     Definition (Group homomorphism). A map $\phi:G\to G'$ is called a group homomorphism […]
  • If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian GroupIf Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$. Then show that $G$ is an abelian group.   Proof. Let $x$ and $y$ be elements of $G$. Then we have \[1=(xy)^2=(xy)(xy).\] Multiplying the equality by $yx$ from the right, we […]
  • A Group Homomorphism and an Abelian GroupA Group Homomorphism and an Abelian Group Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$. Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.   Proof. $(\implies)$ If $G$ is an abelian group, then $f$ […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Abelian Group problems and solutions
A Group is Abelian if and only if Squaring is a Group Homomorphism

Let $G$ be a group and define a map $f:G\to G$ by $f(a)=a^2$ for each $a\in G$. Then prove that...

Close