Property related to the order of a group

Question

Let $a$ and $b$ be elements of a group $G$. Assume that $a$ has order 7 and the $a^3b=ba^3$. Prove that $ab=ba$

Answer 1

Let $A=a^3$, since $a$ has order $7$, you can check that $a=A^5$.

Now, if $A$ and $b$ commute, then $b$ commute with any power of $A$, including $a$.

Answer 2

Hint: There are two steps. Firstly, prove that if $xy=yx$ then $x^iy=yx^i$ for all $i$ ($x$ and $y$ are elements in an arbitrary group). Secondly, find an $i$ such that $a^{3i}=a$.

Answer 3

If $a$ has order $7$, then $a^3$ is a generator of $\langle a \rangle$ since $3$ and $7$ are coprime. So if also $a^3 \in C_G(b)$, it follows that $\langle a \rangle = \langle a^3 \rangle \leq C_G(b)$.