Let $G$ be a finite group and $A:=\{a\in G\mid a\neq a^{-1}\}$. Prove that $|A|$ is even. 
Let $G$ be a finite group and $A:=\{a \in G\mid a \neq a^{-1} \}$ a set that contains all the elements of $G$ that are not equal to their respective inverses. Prove that $A$ contains an even number of elements.

I have seen some posts here here about this proof, but none of which were similar to my attempt.
Here's my attempt:
Since $G$ is finite, then $A$ is also finite.
In addition, every element of $A$ has an inverse because $G$ is a group.
Now, divide $A$ in two sets called $X$ and $Y$, such that $X\subseteq A$ and $Y\subseteq A$, so that every element of $X$ has its inverse in $Y$.
Let $k_{1},k_{2} \in \mathbb{N}$, such that $\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$.
Since there is no element equal to its inverse in $A$, then $ \left | A \right | = \left | X \right | + \left | Y \right |$.
Moreover, $\left | X \right | = \left | Y \right |$ because $A$ only contains elements which are different from their respective inverses.
So,
\begin{aligned}
\left | A \right | &= \left | X \right | + \left | Y \right | \\
&= k_{1} + k_{2} && \text{[$\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$]} \\
&= k_{1} + k_{1} && \text{[$\left | X \right | = \left | Y \right |$]} \\
&= 2\cdot k_{1}
\end{aligned}
$2k_{1}$ is an even number, by the definition of even number.
Therefore, the set $A$ contains an even number of elements.
Does my proof look fine? Every help is appreciated!
 A: There is a simpler division of $A$ that does the trick. Instead of dividing $A$ into two disjoint subsets of equal cardinality, divide it into a pairwise disjoint collection of two element subsets:
$$A = \bigcup_{x \in A} \{x,x^{-1}\}
$$
A: Yuo have a nice idea, but let's highlight some issues:

*

*

Now, divide $A$ in two sets called $X$ and $Y$, such that $X\subseteq A$ and $Y\subseteq A$, so that every element of $X$ has its inverse in $Y$.

Ok, here are a couple of instances: (a) $X=\emptyset$ and $Y=A$; (b) $X=A$ and $Y=A$. In order for the idea to work, you have to add: (i) that, for all $x\in A$, $x\in X$ or $x\in Y$; (ii) that $X\cap Y=\emptyset$


*If you add the conditions above, $\lvert X\rvert=\lvert Y\rvert$ is mainly due to the fact that $x\mapsto x^{-1}$ is an injection from $X$ to $Y$ (which is true regardless of (i) or (ii)), plus surjectivity thanks to (i). Condition (ii) or the fact that no element is equal to its inverse are irrelevant for this end.


*On the other hand, (ii) is needed to prove that $\lvert A\rvert=\lvert X\rvert+\lvert Y\rvert$.
The attentive reader could also question the existence of a pair of sets such as in (1).
