Proving commutativity in a subset $H \subset G$ of a finite group given $x^{-2}y^5 \in H, \forall x, y \in H$ and $x^{-2}y^5 = y^5x^{-2}$ Let $n \geq 1, n \in \mathbb{N}$ and $(G, \cdot)$ a group with $|G| = 10n + 1$ and $H \subset G$ such that $x^{-2}y^5 \in H, \forall x, y \in H$. Prove that if $x^{-2}y^5 = y^5x^{-2}, \forall x, y \in H$, then $xy = yx, \forall x, y \in H$.
I feel like this problem is a dead end for me. I first started by choosing $x \in H$, and by setting $y \rightarrow x$ in the hypothesis, we obtain that $x^3 \in H$ (1). Next, knowing that both $x, x^3 \in H$, I set $x \rightarrow x^3$ and $y \rightarrow x$ in the hypothesis, obtaining $(x^3)^{-2}x^5 = x^{-6}x^5 \in H$, thus $x^{-1} \in H, \forall x \in H$ (2).
After proving that the inverse of any element in $H$ resides in $H$, I had the idea of maybe showing that $H$ is a subgroup of $G$, then we have to somehow prove that $xy \in H$ by writing it in terms of the hypothesis, and by using the fact that $x^{-2}y^5 = y^5x^{-2}$ we might be able to prove $xy = yx$.
I then began asking myself if $e \in H$. I thought about showing $x^{10k + 1} \in H, \forall x \in H, k \in \mathbb{Z}$, so that we can use a consequence of Lagrange's theorem ($x^{ord(G)} = e$), and therefore prove that $e \in H, \forall n \in \mathbb{N}$. Using (1) and (2), we get that $x^{-3} \in H$, then by the hypothesis, we get that $(x^{-3})^{-2}x^5 \in H \Leftrightarrow x^{11} \in H \Leftrightarrow x^{10 * 1 + 1} \in H$. Likewise, I can prove that $x^{21} = (x^{-3})^{-2}(x^3)^5 \in H \Leftrightarrow x^{21} \in H \Leftrightarrow x^{20 * 1 + 1} \in H$. I went on to produce similar results for $x^{31}$ and $x^{41}$, but I cannot seem to prove the general induction case. Maybe I am wrong.
Lastly, I tried taking $x, y \in H$ and setting $x \rightarrow xy$, therefore obtaining that $(xy)^{-2}y^5 \in H$ and trying to use the commutativity property, but to no avail.
Any hints would be very much appreciated on this problem. If anybody has any solution and is willing to share, I would also highly appreciate that. Thanks a lot for taking the time to read through this!
 A: If $H$ is empty then there is nothing to do. So assume $H$ is nonempty.
Let $k=|G|$. The order of $G$ is coprime to $10$. So there exist $a,b\in\mathbb{Z}$ such that $1 = ak+10b$.
In particular, for every $x\in G$, we have
$$x = x^1 = x^{ak+10b} = (x^k)^a (x^{-2})^{-5b},$$
so $x$ is a power of $x^{-2}$.  Similarly,
$$y = y^1 = y^{ak+10b} = (y^k)^a (y^5)^{2b},$$
so $y$ is a power of $y^5$.
Let $x,y\in H$. Then $x^{-2}$ commutes with $y^5$, and hence commutes with every power of $y^5$; in particular, it commutes with $y=(y^5)^{2b}$. Thus, $x^{-2}$ commutes with $y$.
Since $y$ commutes with $x^{-2}$, it commutes with every power of $x^{-2}$. In particular, it commutes with $x = (x^{-2})^{-5b}$. Therefore, $y$ commutes with $x$.
In conclusion, if $x,y\in H$, then $xy=yx$. Note that we are not asking or requiring that this product lie in $H$. In fact, we do not require that $H$ have the starting property (that if $x,y\in H$ then $x^{-2}y^5\in H$); we only require that if $x,y\in H$, then $x^{-2}y^5=y^5x^{-2}$ hold in $G$,
