Do elements $x^2$ and $y$ commute in group $G = \langle x,y \mid x^4, y^{10}, xyx^{-1}y^{-3} \rangle?$ Do elements $x^2$ and $y$ commute in group$$ G =\langle x,y \mid x^4, y^{10}, xyx^{-1}y^{-3} \rangle?$$
Here's what I could get
$$\begin{align}
x^2y& = xxy\\
& = xy^3x \\
&= xyy^2x\\
& = y^3xy^2x \\
&= y^3xyyx \\
& = y^6xyx \\
&= y^9x^2.
\end{align}$$
Then $yx^2y = x^2$.
Now let's assume that $x^2y = yx^2$, then $yx^2y = y^2x^2$ and I get that $y^2 = 1$.  I think that $y^2 = 1$ is bad in this group. And that's where I am stuck.
So what can I do next to prove that it is impossible?
 A: The group $G$ is the quotient of the free group $A = \langle X, Y\rangle$ by the normal subgroup $B$ generated by the elements $X^4, Y^{10}, XYX^{-1}Y^{-3}$. We identify $x, y\in G$ with the images of $X, Y \in A$.
We will show that $G$ is isomorphic to the semidirect product $H = \Bbb Z/10\Bbb Z \rtimes_\phi \Bbb Z/4\Bbb Z$, where $\Bbb Z/4\Bbb Z$ acts on $\Bbb Z/10\Bbb Z$ via the action $\phi:\Bbb Z/4\Bbb Z \rightarrow (\Bbb Z/10\Bbb Z)^\times$ sending $1$ to $3$.
This is mostly tautological. Firstly, we construct a homomorphism $F:A \rightarrow H$ sending $X$ to $(0, 1)\in H$ and $Y$ to $(1, 0) \in H$. It is clear that all the three generators of $B$ are sent to the neutral element of $H$, hence $F$ induces a homomorphism $f$ from the quotient $G$ to $H$.
Secondly, we construct $g: H \rightarrow G$ sending $(1, 0)$ to $Y$ and $(0, 1)$ to $X$. The relations in $G$ exactly ensures that $g$ is well-defined.
Finally, it is obvious that $f$ and $g$ are inverses of each other.
Therefore $G$ is isomorphic to $H$, and we see immediately that $y^2 \neq 1$ in $G$, since its image in $H$ is not the neutral element.
