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.


closed as off-topic by verret, Cesareo, Derek Holt, ancientmathematician, user1729 Feb 5 at 12:16

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  • $\begingroup$ Well, it is generated by $[y,x]$, and you can calculate that... $\endgroup$ – Arturo Magidin Feb 4 at 17:57
  • 3
    $\begingroup$ (1) The elements of the derived subgroup are not numbers. (2) If $X$ is a subset of $G$ that generates the group, then the normal closure of $\{ [x,y]\mid x,y\in X\}$ is equal to $[G,G]$. Here, $X=\{x,y\}$. $\endgroup$ – Arturo Magidin Feb 4 at 18:15
  • $\begingroup$ @AfsaneBahri You're question has been closed because "Honestly, I do not have any idea" isn't helpful! However, I think if you added in your issue with generating every element of the derived subgroup then it would be fine (replace the word "number" with the word "elements" in your above comment, and write a bit more about Arturo's hint). $\endgroup$ – user1729 Feb 5 at 12:18

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.

  • 1
    $\begingroup$ You use "obviously" and "clearly" too much. For example, it is not clear that this group is non-abelian. To prove that it is non-abelian requires one to prove something about the relationship between $1+p^{n-2}$ and $p^{n-1}$. $\endgroup$ – user1729 Feb 5 at 12:14
  • $\begingroup$ @user1729, sorry for skipping this detail of the proof. Suppose, $M_n(p)$ is abelian. Then it would mean, that $y = y^{p^{n - 2} + 1}$, from which it follows, that the order of $y$ divides $p^{n - 2}$. But $y$ has order $p^{n-1}$, thus we receive a contradiction. $\endgroup$ – Yanior Weg Feb 5 at 12:51

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