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.
Torsion 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 […]
Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.
Then determine the number of elements in $G$ of order $3$.
Proof.
Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.
By […]
The 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 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 […]
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 […]
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, […]
Group 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 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 […]