Which of the following holds? Problem in group theory. Let $G$ be a group whose presentation is $G=\\{x,y\mid x^5=y^2=e, x^2y=yx\\}$ Then $G$ is isomorphic to 1) $\mathbb Z_5$ 2) $\mathbb Z_{10}$ 3) $\mathbb Z_2$ 4) $\mathbb Z_{30}$ I thought that clealy it is only isomorphic to $\mathbb Z_{10}$. But then i am confused with the given relator? Where i am wrong?

Captain Lama's answer is obviously much more comprehensive. But if you want to know how to game the multiple-choice system, read on!

If we're convinced that _one_ of the choices is correct, we can assume that $x$ and $y$ commute. We can do that because all choices are Abelian, hence $xy = yx$ no matter which the group ends up being.

But this means that, since $x^2y = yx$, we have

\begin{align*} x^2y &= yx \tag{given}\\\ x^2y &= xy \tag{commutativity}\\\ x^2 &= x \tag{$y^{-1}$ on the right}\\\ x &=1. \tag{multiply by $x^{-1}$} \end{align*}

Thus we can safely throw $x$ out of the generating set, as well as any relation involving $x$, in this case. Then your group is $\langle y \mid y^2 = e\rangle$, clearly $\Bbb Z_2$.