TITLE: Proving the given identity in group
QUESTION [15 upvotes]: Please consider the following problem:
If in a group G , $xy^2 = y^3x$ and $yx^2 = x^3y$ then prove that $x = y = e$ where $e$ is the identity element.
Here is my attempt so far:
Since it is difficult to extract $x$ or $y$ from either of the equations I am trying to find $xy$ from both the equations and equating them.
$xy^2 = y^3x \\ xy = y^3xy^{-1}  ----- (1) $
$x^3y = yx^2 \\ xy = x^{-2}yx^2 ---- (2)$
Now equating (1) and (2) :
$y^3xy^{-1} = x^{-2}yx^2$
After this step I have filled pages (difficult to reproduce here) substituting variables here and there but somehow I am caught in a loop.
Is there a way out?

REPLY [2 votes]: Lemma 1: $y = x^{3k} y x^{-2k}$ (Induction using $y = x^3 y x^{-2}$)
Lemma 2: $y^n = x^{3^n} y^n x^{-2^n}$ (Induction using Lemma 1)
Lemma 3: $y^n = x^{3^n k} y^n x^{-2^n k}$ (Analogous to Lemma 1)
From Lemma 3 and $y^3 = x y^2 x^{-1}$, we get
$$x^{27m} y^3 x^{-8m} = x^{9n + 1} y^2 x^{-4n - 1}$$
By substituting $y^3 = x y^2 x^{-1}$ we get
$$x^{27m + 1} y^2 x^{-8m - 1} = x^{9n + 1} y^2 x^{-4n - 1}$$
By choosing $m = 1, n = 3$, we get
$$x^{28} y^2 x^{-9} = x^{28} y^2 x^{-13}$$
$$x^4 = e$$
Thus, the order of $x$ divides 4. Since $x^3$ and $x^2$ are conjugates, they must
have the same order. But then $x$ must be of order 1, i.e., $x = e$.
And thus $y = e$ as well.