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 $c+1$.
1Can 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
abstract-algebra group-theory finite-groups computational-mathematics p-groups
edited Dec 27 '18 at 1:01
the_fox
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
add a comment |
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 $c+1$.
1Can 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
abstract-algebra group-theory finite-groups computational-mathematics p-groups
edited Dec 27 '18 at 1:01
the_fox
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
add a comment |
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 $c+1$.
1Can 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
abstract-algebra group-theory finite-groups computational-mathematics p-groups
edited Dec 27 '18 at 1:01
the_fox
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
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 $c+1$.
1Can 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
abstract-algebra group-theory finite-groups computational-mathematics p-groups
abstract-algebra group-theory finite-groups computational-mathematics p-groups
edited Dec 27 '18 at 1:01
the_fox
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
edited Dec 27 '18 at 1:01
the_fox
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
edited Dec 27 '18 at 1:01
the_fox
2,43411431
edited Dec 27 '18 at 1:01
the_fox
2,43411431
edited Dec 27 '18 at 1:01
the_fox
2,43411431
2,43411431
asked Dec 3 '18 at 13:13
A.Messab
507
asked Dec 3 '18 at 13:13
A.Messab
507
asked Dec 3 '18 at 13:13
A.Messab
507
507
add a comment |
add a comment |
1 Answer
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$P_1(G) = [G,G]G^p$. 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.
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
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
1
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
add a comment |
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$P_1(G) = [G,G]G^p$. 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.
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
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
1
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
add a comment |
$P_1(G) = [G,G]G^p$. 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.
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
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
1
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
add a comment |
$P_1(G) = [G,G]G^p$. 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.
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
$P_1(G) = [G,G]G^p$. 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.
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
answered Dec 3 '18 at 17:31
Derek Holt
52.6k53570
52.6k53570
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
1
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
add a comment |
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
1
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
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
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
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
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
1
1
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
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
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
Many many thanks professor, that is so helpful demonstration
– A.Messab
Dec 4 '18 at 11:09
add a comment |
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