Mixing
How to proof that $G$ and $G^*$ has the same number of generators?
Let $G$ be a $d-$generator $p-$group, Let $F$ be the free group of rank $d$ freely generated by $a_1$ . . . . . $a_d$, and let $R$ be the
kernel of a homomorphism $theta$ from $F$ onto $G$; Define $R^*$ to be $[R, F]R^p$ and $G^*$ to be $F/R^*$.
Does $G$ and $G^*$ has the same number of generators?
abstract-algebra finite-groups
asked Nov 21 '18 at 9:19
A.Messab
507
add a comment |
Let $G$ be a $d-$generator $p-$group, Let $F$ be the free group of rank $d$ freely generated by $a_1$ . . . . . $a_d$, and let $R$ be the
kernel of a homomorphism $theta$ from $F$ onto $G$; Define $R^*$ to be $[R, F]R^p$ and $G^*$ to be $F/R^*$.
Does $G$ and $G^*$ has the same number of generators?
abstract-algebra finite-groups
asked Nov 21 '18 at 9:19
A.Messab
507
add a comment |
Let $G$ be a $d-$generator $p-$group, Let $F$ be the free group of rank $d$ freely generated by $a_1$ . . . . . $a_d$, and let $R$ be the
kernel of a homomorphism $theta$ from $F$ onto $G$; Define $R^*$ to be $[R, F]R^p$ and $G^*$ to be $F/R^*$.
Does $G$ and $G^*$ has the same number of generators?
abstract-algebra finite-groups
asked Nov 21 '18 at 9:19
A.Messab
507
Let $G$ be a $d-$generator $p-$group, Let $F$ be the free group of rank $d$ freely generated by $a_1$ . . . . . $a_d$, and let $R$ be the
kernel of a homomorphism $theta$ from $F$ onto $G$; Define $R^*$ to be $[R, F]R^p$ and $G^*$ to be $F/R^*$.
Does $G$ and $G^*$ has the same number of generators?
abstract-algebra finite-groups
abstract-algebra finite-groups
asked Nov 21 '18 at 9:19
A.Messab
507
asked Nov 21 '18 at 9:19
A.Messab
507
asked Nov 21 '18 at 9:19
A.Messab
507
asked Nov 21 '18 at 9:19
A.Messab
507
asked Nov 21 '18 at 9:19
A.Messab
507
507
add a comment |
add a comment |
1 Answer
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In the following I will assume that $theta: F to G$ is not just any homomorphism of groups but a surjective one, so that we have an isomorphism $F/R cong G$ (or a presentation $langle a_1, dots, a_d:|: R rangle$)
Similarly, $G^*$ admits - by definition - a surjective morphism $F to G^*$ or the presentation $langle a_1, dots, a_d:|: [R,F]R^q rangle$. In particular, $G^*$ is generated by $d$ elements.
It might also be noteworthy that if $G$ is not generated by less than $d$ elements, then this is true for $G^*$, too, since $G$ is a quotient of $G^*$.
Also note that the number of generators is not something which is well-defined per se as any element of a group may participate in a generating set and minimal generating sets might not have the same size (for example $Bbb{Z}$ has minimal generating sets ${1}$ and ${2,3}$). You can ask if the sizes of smallest generating sets of $G$ and $G^*$ is identical and the above considerations show that this is actually the case.
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
add a comment |
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In the following I will assume that $theta: F to G$ is not just any homomorphism of groups but a surjective one, so that we have an isomorphism $F/R cong G$ (or a presentation $langle a_1, dots, a_d:|: R rangle$)
Similarly, $G^*$ admits - by definition - a surjective morphism $F to G^*$ or the presentation $langle a_1, dots, a_d:|: [R,F]R^q rangle$. In particular, $G^*$ is generated by $d$ elements.
It might also be noteworthy that if $G$ is not generated by less than $d$ elements, then this is true for $G^*$, too, since $G$ is a quotient of $G^*$.
Also note that the number of generators is not something which is well-defined per se as any element of a group may participate in a generating set and minimal generating sets might not have the same size (for example $Bbb{Z}$ has minimal generating sets ${1}$ and ${2,3}$). You can ask if the sizes of smallest generating sets of $G$ and $G^*$ is identical and the above considerations show that this is actually the case.
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
add a comment |
In the following I will assume that $theta: F to G$ is not just any homomorphism of groups but a surjective one, so that we have an isomorphism $F/R cong G$ (or a presentation $langle a_1, dots, a_d:|: R rangle$)
Similarly, $G^*$ admits - by definition - a surjective morphism $F to G^*$ or the presentation $langle a_1, dots, a_d:|: [R,F]R^q rangle$. In particular, $G^*$ is generated by $d$ elements.
It might also be noteworthy that if $G$ is not generated by less than $d$ elements, then this is true for $G^*$, too, since $G$ is a quotient of $G^*$.
Also note that the number of generators is not something which is well-defined per se as any element of a group may participate in a generating set and minimal generating sets might not have the same size (for example $Bbb{Z}$ has minimal generating sets ${1}$ and ${2,3}$). You can ask if the sizes of smallest generating sets of $G$ and $G^*$ is identical and the above considerations show that this is actually the case.
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
add a comment |
In the following I will assume that $theta: F to G$ is not just any homomorphism of groups but a surjective one, so that we have an isomorphism $F/R cong G$ (or a presentation $langle a_1, dots, a_d:|: R rangle$)
Similarly, $G^*$ admits - by definition - a surjective morphism $F to G^*$ or the presentation $langle a_1, dots, a_d:|: [R,F]R^q rangle$. In particular, $G^*$ is generated by $d$ elements.
It might also be noteworthy that if $G$ is not generated by less than $d$ elements, then this is true for $G^*$, too, since $G$ is a quotient of $G^*$.
Also note that the number of generators is not something which is well-defined per se as any element of a group may participate in a generating set and minimal generating sets might not have the same size (for example $Bbb{Z}$ has minimal generating sets ${1}$ and ${2,3}$). You can ask if the sizes of smallest generating sets of $G$ and $G^*$ is identical and the above considerations show that this is actually the case.
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
In the following I will assume that $theta: F to G$ is not just any homomorphism of groups but a surjective one, so that we have an isomorphism $F/R cong G$ (or a presentation $langle a_1, dots, a_d:|: R rangle$)
Similarly, $G^*$ admits - by definition - a surjective morphism $F to G^*$ or the presentation $langle a_1, dots, a_d:|: [R,F]R^q rangle$. In particular, $G^*$ is generated by $d$ elements.
It might also be noteworthy that if $G$ is not generated by less than $d$ elements, then this is true for $G^*$, too, since $G$ is a quotient of $G^*$.
Also note that the number of generators is not something which is well-defined per se as any element of a group may participate in a generating set and minimal generating sets might not have the same size (for example $Bbb{Z}$ has minimal generating sets ${1}$ and ${2,3}$). You can ask if the sizes of smallest generating sets of $G$ and $G^*$ is identical and the above considerations show that this is actually the case.
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
answered Nov 21 '18 at 9:34
Matthias Klupsch
6,1591227
6,1591227
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
add a comment |
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
Many thanks for your answer, the fact is I understand that you used isomorphism theorem to establish those tow isomorphism relations! Yet, how did you commute between the assertion of isomorphic to the presentation?
– A.Messab
Nov 21 '18 at 9:51
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
What is a presentation of a group? Writing $G = langle a_1, a_2, dots, a_d:|: Rrangle$ is just another way of saying that there exists a surjective morphism $F to G$ from the free group on $d$ generators and the kernel of this morphism is the smallest normal subgroup of $F$ containing $R$ (which is $R$ if $R$ is itself a normal subgroup). As an example you might consider what writing $D_{8} = langle s,t :|: s^2, t^4, sts^{-1}t rangle$ actually means.
– Matthias Klupsch
Nov 21 '18 at 10:12
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
Many many many thanks; I was reading polycyclic presentations to get a deep understanding for what so-called p-generating, your definition for presentation is the most "meaningful" that I "encounter" with. My best regards
– A.Messab
Nov 21 '18 at 10:20
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
I am glad that I could help you. If you are satisfied with my answer, you might consider accepting it.
– Matthias Klupsch
Nov 21 '18 at 10:23
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
Sorry to didn´t that from the first, I was so excited!
– A.Messab
Nov 21 '18 at 10:27
add a comment |
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