Why is the $p-$multiplicator of a $p-$group is an elementary abelian $p -$group?
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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.
abstract-algebra p-groups computational-algebra
asked yesterday
A.Messab
184
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up vote
1
down vote
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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.
abstract-algebra p-groups computational-algebra
asked yesterday
A.Messab
184
1
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 $p$.
– Tobias Kildetoft
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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.
abstract-algebra p-groups computational-algebra
asked yesterday
A.Messab
184
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.
abstract-algebra p-groups computational-algebra
abstract-algebra p-groups computational-algebra
asked yesterday
A.Messab
184
asked yesterday
A.Messab
184
asked yesterday
A.Messab
184
asked yesterday
A.Messab
184
asked yesterday
A.Messab
184
184
1
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 $p$.
– Tobias Kildetoft
yesterday
add a comment |
1
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 $p$.
– Tobias Kildetoft
yesterday
1
1
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 $p$.
– Tobias Kildetoft
yesterday
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 $p$.
– Tobias Kildetoft
yesterday
add a comment |
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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 $p$.
– Tobias Kildetoft
yesterday