Derived subgroup of $\langle{x,y\,|\, x^p=y^{p^{n-1}}=1,\,{{x^{-1}}{yx}}={y^{1+p^{n-2}}}}\rangle$.

Question

I would like to prove that if $M_n(p)=\langle{x,y\,|\, x^p=y^{p^{n-1}}=1,\,{{x^{-1}}{yx}}={y^{1+p^{n-2}}}}\rangle$, then $M'_n(p)$ is a cyclic group of order $p$.

I was wondering if someone could help me. Honestly, I do not have any idea.

Answer 1

Your group can be written in the form $M_n(p) = C_{p^{n-1}} \rtimes C_{p}$, where the semidirect product is taken with respect to homomorphism $\phi: C_p \rightarrow \langle \psi: y \mapsto y^{1 + p^{n - 2}} \rangle$ (You can find more information on semidirect products here: https://en.wikipedia.org/wiki/Semidirect_product) This group is obviously metabelian. Thus, its derived subgroup is abelian. Your group is clearly non-abelian, thus the derived subgroup is non-trivial. Now, you can see, that $\frac{M_n(p)}{\langle y^{p^{n - 2}} \rangle}$ is abelian and has order $p^{n-1}$. Thus the order of the derived subgroup is at most $\frac{|M_n(p)|}{|\frac{M_n(p)}{\langle y^{p^{n - 2}} \rangle}|} = \frac{p^{n}}{p^{n - 1}} = p$. Thus it is $p$ as $M_n'(p)$ is a non-trivial subgroup of a $p$-group. That means $M_n'(p) \cong C_p$ Q.E.D.