If $|a^{2}|=|b^{2}|$ then $|a|=|b|?$

Question

If $|a^{2}|=|b^{2}|$ (for non identity elements $a$ and $b$ of a group $G$ and $|a|$ denotes the order of the element $a$) prove or disprove that $|a|=|b|.$

I tried as follows

Clearly infinite order case is trivial. Let is discuss finite order case $$|a^{2}|=|b^{2}|$$ $\Rightarrow$$$\frac{|a|}{(2,|a|)}=\frac{|b|}{(2,|b|)} $$ where $(2,|a|)$ is the $\gcd$. Now I am stuck. How proceed further. Thanks in advance.

Answer 1

Hint: you have already reduced this to a question about elementary number theory: are there two numbers $m,n > 1$ such that $m \ne n$ but $m/(2,m) = n/(2,n)$?

Answer 2

(EDIT along Erick Wong's comment - I overlooked the (somewhat irrelevant) condition that the elements should not be the identity).

Consider the cyclic group of order $6$. Let $a$ be a generator and $b=a^2$. Then $a^2=b$ and $b^2$ have order $3$, but $a$ has order $6$.