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?
 A: If you square both sides, you get $a^6=b^6$.  But $a^5=e=b^5$, so $a=b$.
To comment on your solution, note that $o(2)=\infty$ is impossible in a finite group.  In $\mathbb{Z}_{10}$ (which always implies the additive structure), the order of $2$ is $5$ (nice and finite).
A: Since $$b^5=a^5 =a^3a^2 =b^3a^2\implies a^2=b^2$$
so $$b^3=a^3 =a^2a =b^2a\implies a=b$$
A: No, it is true, the special case $\,m,n = 3,5\,$ below.
Lemma $\ \gcd(m,n) = 1,\ a^m = b^m,\ a^n = b^n\,\Rightarrow\, a = b$
Proof $\ $ The set $\,S\,$ of $\,k\in \Bbb Z\,$ such that $\,a^k = b^k\,$ is closed under subtraction hence, by  a 1-line proof, its least positive element $\,d\,$ divides every element, so $\,d\mid m,n\,$ coprime, thus $\,d =1.$
A: As others have shown, $a^3=b^3$ implies $a=b,$ where $a$ and $b$ are group elements with order $5.$ Your example of $2$ and $8$ representing integers modulo $10$ does not disprove this theorem.  I'm not sure whether you were talking about the integers modulo $10$ under addition or multiplication (the latter is not a group!), but $2+2+2 \not  \equiv 8+8+8 \pmod {10}$ and $2 \times 2 \times 2 \not \equiv 8 \times 8 \times 8 \pmod {10}$, so the hypothesis is not met and it is not a problem that $2 \not \equiv 8;$ the theorem is not disproved.
