Here's a problem I have been thinking about all day.
Let $H$ be a subgroup with $h$ elements of a group $G$. Suppose that $G$ has an element $a$ such that for all $x$ in $H$, $(xa)^3 = 1$, the identity. In $G$, let $P$ be the set of all products $x_1ax_2a \ldots x_n a$, with $n$ a positive integer and the $x_i$ in $H$. Show that $P$ has no more than $3h^2$ elements.
Here's is my attempt:
By setting $x=1$, we learn that $a^3=1$ and so $a=1$ or $\mathrm{ord}(a)=3$. Now if $a\in H$ then we are done since, $P\subset H$ whose order is at most $h$.
Hence, assume not and $\mathrm{ord}(a)=3$. Now we notice that $x_nax_na,\ldots,x_2ax_2a x_1ax_1a\in P$ is the inverse of $x_1ax_2a,\ldots,x_na$ and $1a1a1a=1\in P$. Therefore, $P$ is a subgroup of $G$.
At this point, I was hoping to use Lagrange's theorem to bound the order of $P$. I have tried many approaches but none of them led me anywhere meaningful.
I am hoping to get some hints. I prefer hints to complete solutions. Thank you for your time.
Since you want hints, this is not a complete solution.
Try to prove that every element of $P$ can be written as $xay$, $xa^{-1}y$, or $xa^{-1}ya$, with $x,y \in H$, thereby giving at most $3h^2$ elements.
Here is one step in the proof, to give you an idea.
Let $x,y,z \in H$. Then, using $(yza)^3=1$, we get $(xay)(za) = x(yz)^{-1}a^{-1}(yz)^{-1}$.
Incidentally, arguments like this are used in the proof that finitely generated groups of exponent 3 are finite.