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.

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(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]

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Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.
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By […]

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Let $G$ be a finite group. Let $a, b$ be elements of $G$.
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Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
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Problem.
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Lagrange's Theorem
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Definition/Hint
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