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


$P_1(G) = [G,G]G^p$. 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.

  • $\begingroup$ 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

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