Isomorphism of a group $G = \left\langle x , y \mid x^5 = y^2 = e , x^2y = yx \right\rangle$ Let $G$ be a group whose representation is $G = \left\langle x , y \mid x^5 = y^2 = e , x^2y = yx \right\rangle$ . Then $G$ is isomorphic to
a) $\mathbb Z_5$
b) $\mathbb Z_2$
c) $\mathbb Z_{10}$
d) $\mathbb Z_{30}$
I think order of $G$ is 10, but how to show that $G$ is cyclic. 
any help would be appreciated, Thank you.  
 A: If the question is  $\{x,y:x^5=y^2=e, x^2 y=yx\}$ then answer is (b) i.e $z_2$ as $(x^2y)^2$ will give $x=e$ (use $x^2 y=yx$) so only y will remain and thus order of group is two.
A: I'm working under the assumption that the OP meant $G=\left\langle x , y \mid x^5 = y^2 = e , x^2y = yx \right\rangle$.  (He wrote $G=\left\langle x , y \mid x^5 = x^2 = e , x^2y = yx \right\rangle$, which makes the problem very trivial as it means $G=\langle y\rangle$ without restriction on $y$, so $G\cong\mathbb{Z}$.)
The subgroup $N:=\langle x\rangle$ is normal in $G$ of order $1$ or $5$.  The factor group $G/N$ is generated by $yN$, whence its order is $1$ or $2$.  Consequently, we have $|G|= |G/N|\cdot |N|$ is a divisor of $2\cdot 5=10$.
Now, if $N$ is of order $5$, then it follows that $G/N$ must be of order $2$ (otherwise, the relation $x^2y=yx$ would imply that $x^2=x$, which is a contradiction).  Consider the element $yx$.  We have $(yx)^2=yxyx=x^2yyx=x^3$.  Therefore, $(yx)^{10}=x^{15}=e$.  Note that $yx\neq e$ (as $y\notin N$).  If $(yx)^5=e$, then $\langle yx\rangle\neq N$ is another subgroup of $G$ of order $5$, but this contradicts the Second Sylow Theorem (alternatively, you can see that $N\cap\langle yx\rangle$ contains $x^3\neq e$, but as $N\cong\mathbb{Z}_5$, we must have $\langle yx\rangle=N$, a contradiction).  If $(yx)^2=e$, then $x^3=e$, which contradicts the assumption that $N$ is of order $5$.  Therefore, the order of $yx$ is exactly $10$.  This means $G=\langle yx\rangle\cong\mathbb{Z}_{10}$, so $G$ is abelian.  Ergo, we have $xy=yx=x^2y$, leading to $x=e$, which is absurd.  Hence, $N$ is the trivial subgroup.
We have established that $x=e$.  That is, $G=\left\langle y\,|\,y^2=e\right\rangle\cong\mathbb{Z}_2$.
