About quotients of Lower Exponent$-p$ central Series

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 $$i, is $$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 $$Q$$.
• 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. – A.Messab Dec 4 '18 at 9:19
• An abelian group $G$ in which, for some prime $p$, $g^p=1$ for all $g \in G$ is an elementary abelian $p$-group. – Derek Holt Dec 4 '18 at 10:08
• know the definition, but why if it is a quotient of some group containing $G^p$ it is elementary abelian, and of exponent $p$? – A.Messab Dec 4 '18 at 10:14
• 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. – Derek Holt Dec 4 '18 at 11:02