Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$

Question

Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.

I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used this to obtain $yx^{-1} = x^{-p-1} y$, which will allow me to switch $y$ and $x^{-1}$ in each $yx^{-1}$. However, I still cannot figure out where to go from here. Any help is appreciated.

Answer 1

$yx^k = x^{k(p+1)}y$, so $y^a x^k = x^{k(p+1)^a} y^a$, so:

$$\begin{array}{rcl} (yx^{-1})^p &=& y x^{-1} y x^{-1} y x^{-1} \cdots yx^{-1} \\ &=& x^{-(p+1)}y^2 x^{-1} y x^{-1} \cdots yx^{-1} \\ &=& x^{-(p+1)-(p+1)^2} y^3 x^{-1} \cdots yx^{-1} \\ &=& \cdots \\ &=& x^{-(p+1)-(p+1)^2-\cdots-(p+1)^{p-1}} y^p \\ &=& x^{-(p+1)-(2p+1)-\cdots-((p-1)p+1)} x^p \\ &=& x^{-\frac{(p-1)p}2 p - p} x^p \\ &=& x^{-\frac{(p-1)}2 p^2} \\ &=& 1 \\ \end{array}$$

where $x^{(p+1)^n} = x^{1+np+O(p^2)} = x^{1+np}$ because $x^{p^2}=1$.

Answer 2

From $yxy^{-1} = x^{p+1}$ it follows that $yx^{-1}y^{-1} = (x^{p+1})^{-1}$, so $yx^{-1} = x^{-p-1}y$. We can now prove by an easy induction on $k$ that $$ (yx^{-1})^k = x^{-p(1+2+\ldots+k)-k}y^k. $$ For $k=p$, we obtain $$ (yx^{-1})^p = x^{-p(1+2+\ldots+p)-p}y^p = x^{-p}y^p = 1 $$ since $p(1+2+\ldots+p) = p \cdot \frac{p(p+1)}{2}$ is a multiple of $p^2$.