Let $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$. 
Prove or disprove: let $G$ be a group and $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$.

I saw the following example which tries to disprove the theorem: $G=\mathbb{Z}_{10}$ and $a=2,b=8$.
I'm not sure about that part, but $o(2)=5$ and $o(2)=8$. I think that $o(2)=\infty$ because there is not $n$ so $2^n\,mod\,10=1$ but I'm not sure.
Does this example disproves the theorem?

Since $$b^5=a^5 =a^3a^2 =b^3a^2\implies a^2=b^2$$
so $$b^3=a^3 =a^2a =b^2a\implies a=b$$
