Uniqueness of element Group element $a \in G$ have order 6 where $G$ can be any group. Prove that  $a^3, a^4$ are the only elements $b,c$ in the group $G$ such that $a = bc$, $|b| = 2$, $|c| = 3$ and $bc = cb$.
Because $G$ can be any group, I'm really not sure how this could stand. 

Hint: Suppose that $b$ and $c$ are any such elements satisfying those assumptions. Squaring both sides of $a=bc$ yields $a^2=b^2c^2=c^2$. (This is where it is important that $b$ and $c$ commute.) 
Use this to solve $c$ in terms of $a$, and that will then give you $b$ in terms of $a$. Mouse over below if you get stuck.

 Since $c$ has order $3$, $a^2=c^2=c^{-1}$, so $c=a^{-2}=a^4$, since $a$ has order $6$. Then $a=ba^4$, so $b=a^{-3}=a^3$. You should also check that $a^3$ and $a^4$ have the claimed properties, but that's almost immediate.

