2
$\begingroup$

Let $G$ be a finite $p-$group of number of generators $d$ and exponent$-p$ class $c$, that is $c$ is the smallest integer satisfying $P_c(G) =1$ in the series
$$ G=P_0(G) \geq ...\geq P_{i-1}(G)\geq P_{i}(G)\geq ... $$ Where $P_{i}(G)= [P_{i-1}(G), G]{P_{i-1}(G)}^p$.

A group $H$ is a descendant of $G$ if $H$ has generator number $d$ and the quotient $H/P_c(H)$ is isomorphic to $G$

A group is an immediate descendant of $G$ if it is a descendant of $G$ and has an exponent$-p$ class $c+1$.

1\Can you show me "in details" why $G/P_1(G)$ is an elementary abelian? What is its order then?

2\ Is $G$ a descendant of $G/P_1(G)$?

3\ For $i<c$, is $G/P_{i+1}(G)$ is an "immediate descendant" of $G/P_i(G)$?

My best regards

$\endgroup$
4
$\begingroup$

$P_1(G) = [G,G]G^p$. Then $G/P_1(G)$ is abelian because $[G,G] \le P_1(G)$. It has exponent $p$, and is therefore elementary abelian because $G^p \le P_1(G)$.

The answers to 2 and 3 are both yes. Let $Q = G/P_1(G)$. Then, since $P_1(G) = \Phi(G)$, the generator numbers of $Q$ and $G$ are the same. Since $Q$ has exponent $p$-class $1$, it follows that $G$ is a descendant of $Q$.

3 also follows directly from the definition of immediate descendant.

$\endgroup$
  • $\begingroup$ Thank you very much for your answer, I understood every thing but the fact that "it has exponent $p$ and therefore elementary abelian because $G^p \geq P_1(G)$", please some explanation. $\endgroup$ – A.Messab Dec 4 '18 at 9:19
  • $\begingroup$ An abelian group $G$ in which, for some prime $p$, $g^p=1$ for all $g \in G$ is an elementary abelian $p$-group. $\endgroup$ – Derek Holt Dec 4 '18 at 10:08
  • $\begingroup$ know the definition, but why if it is a quotient of some group containing $G^p$ it is elementary abelian, and of exponent $p$? $\endgroup$ – A.Messab Dec 4 '18 at 10:14
  • 1
    $\begingroup$ Let $K=P_1(G) = [G,G]G^p$ and $Q=G/K$. Then, for any $gK,hK \in Q$, $[gK,hK]=[g,h]K = K$, since $[G,G] \le K$, so $Q$ is abelian. Also, for any $gK \in Q$, $(gK)^p = g^pK = K$ because $g^p \in G^p \le K$. So $Q$ is elementary abelian. $\endgroup$ – Derek Holt Dec 4 '18 at 11:02
  • $\begingroup$ Many many thanks professor, that is so helpful demonstration $\endgroup$ – A.Messab Dec 4 '18 at 11:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.