The formal question is that in a group $G$, with finite elements and an element $a$, which is a non-identity element, why is it true that $|a| = 2|a^2|$?
I thought of it as a similar proof to that of $|a| = |a^{-1}|$, and tried to go from there as well as use that fact.
My proof so far:
Attempting to prove my contradiction. Suppose $|a| = n$, where $n$ is the smallest positive integer such that $|a^n = e|$, $e$ being the identity element.
$$2(a^2)^n = > 2(a^2)^ne=2(a^2)^na^n=2(aaa...a)(aaa...a)(a^{-1}a^{-1}a^{-1}...a^{-1})_{n-times}$$
$$= 2(a^n)$$ and we also know that $(a^{-1})^n = e$
From there, we suppose that there exists a $K$ in the set of positive
integers, such that $(a^{-1})^k = e$ and $k < n$.
$$a^k = a^ke$$ and so on to finish the proof.
I'm stuck from here, what is the next step I can make to progress the proof? How can I show that 2(a^2)^n is equal to $e$ so I can prove by contradiction?
This is not true. Take any group containing an element with odd order - the cyclic group with $5$ elements for example. Any non-identity element $a$ will have an odd order by Lagrange's Theorem and hence $|a| \neq 2|a^2|$, since this would imply $a$ to be even.
Edit: In the case the element $a$ has even order this is true. Then, if $|a^2| < \frac{|a|}{2}$ we have a contradiction - that $|a|$ is not the order of $a$, i.e there exists $n < |a|$ such that $a^n = 1$. Clearly $(a^2)^\frac{|a|}{2} =1$ and so we deduce the result.
Second Edit: This is not necessarily true if your group has even order, consider the cyclic group with $6$ elements $C_6 = \{1,g,g^2,...,g^5\}$. Let $a = g^2$ then $|a|=|g^2|=3$ and $|a^2|=|g^4| = 3$. We need to assume $a$ has even order!