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Let $G$ be a $p-$group that have the finite presentation $F/R$($F$ is a free group of rank $d$); The $p-$multiplicator of $G$ is defined by $$ G^* = R/[F,R]R^p $$ Why $G^*$ in an elementary abelian $p -$group?

Many thanks.

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    $\begingroup$ This is directly from the way it is defined. You have $[R,R]\subseteq [F,R]$ so it is abelian, and you have included $R^p$ so it has exponent $p$. $\endgroup$ – Tobias Kildetoft Nov 30 '18 at 14:54
  • $\begingroup$ @TobiasKildetoft Thank you very much for your answer. $\endgroup$ – A.Messab Dec 3 '18 at 12:12

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