Let $p$ be an odd prime and $G$ be a group generated by elements $x,y$. Given $$x^{p^{2}}=1, x^{p} = y^{p}, yxy^{-1} = x^{p+1}, (yx^{-1})^{p} = 1,$$ show that $G$ is the semidirect product of a two cyclic subgroups of order $p^{2}$ and $p$, respectively.
The given conditions imply that $|x| = p^{2}, |yx^{-1}|=p$. I think we can pick $x$ and $yx^{-1}$ to be the generator of $\mathbb{Z}/p^{2}\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$, respectively. However, it isn't clear to me where this is leading. Any help is appreciated.
Hint: Try to show that it is the split extension of $\Bbb Z_p$ by $\Bbb Z_{p^2}$ with the homomorphism $\varphi:\Bbb Z_p\hookrightarrow\Bbb Z_{p^2-p}\cong\rm{Aut}(\Bbb Z_{p^2})$ given by the inclusion.