an example of a group $G$ such that it has elements $a,b$ of order $6$ and $2$ respectively satisfying $a^3b = ba$ Can anybody give me an example of a group $G$ such that it has elements $a,b$ of order $6$ and $2$ respectively satisfying $a^3b = ba$
I came across a problem saying - "$G$ is a group. $a,b$ are two elements of $G$ such that $o(a)=6 ,o(b)=2$ and $a^3b =ba$. Then find order of $ab$." But if $a,b$ satisfy the above properties then it's eventually turning out that $a^2 = e$. 
 \begin{align*}a^3b &=ba\\
  b = a^3ba &= ba^2\\
  a^2 &= e
\end{align*}
So I think that the question is wrong. I think there does not exist any group satisfying such properties. Can anybody give me some proper justification behind this ? Or have I done any mistake ? 
 A: Indeed, such a group cannot exist (e.g. by the argument you use). A formal way to see this (though it uses tools that you might not yet have used) is to consider the formal presentation
$
\langle a,b\mid a^6=b^2=1,a^3b=ba\rangle
$
and to use a tool called coset enumeration which verifies that under these conditions the group would have order $4$ and the order of $a$ (which must divide $6$) would be $2$ only.
A: You are entirely correct; if $o(a)=6$ and $o(b)=2$ and $a^3b=ba$ then
$$b=a^6b=a^3(a^3b)=a^3(ba)=(a^3b)a=(ba)a=ba^2,$$
and hence $a^2=e$, contradicting the fact that $o(a)=6$. Hence no such group exists.
Alternatively, note that the given relation is equivalent to $a^3=bab^{-1}$, which implies
$$e=a^6=(a^3)^2=(bab^{-1})^2=ba^2b^{-1},$$
and hence that $a^2=e$, again contradicting $o(a)=6$.
On the other hand, if we replace the relation $a^3b=ba$ by $a^5b=ba$ then we get a familiar group, and the exercise makes sense.
A: $a^3 = bab^{-1}$ and so $a = b^2ab^{-2} = (a^3)^3 = a^9$. Thus $a^8 = 1$, a contradiction to $o(a) = 6$.
