Relation between Minimum number of generator and number of elements?
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For a finite group that needs n generator to construct, is the number of elements of the group the same as the product of the numbers of elements each of these generators alone can generate?
If so,is it also true that for two finite groups, if they have the same minimum number of generators needed to construct and that each generator alone can generate the same number of elements corresponding to another group, then the two groups are isomorphic?
For example, if I two groups that need 2 generators,and one of the generator generates 3 elements(including e) and another one generates 6 elements, then (3-1)(6-1)=10 is the number of elements of the group and the two groups are isomorphic to each other?
I just started learning group theory a few days ago, so hope you don't mind if I actually ask something stupid.
Briefly , My attempt is that the set generated by a generator is a cyclic group, and that the direct product of those cyclic groups should be isomorphic to the original group itself.
group-theory
asked Sep 13 at 0:19
Johnny Chen
1026
add a comment |Â
up vote
3
down vote
favorite
For a finite group that needs n generator to construct, is the number of elements of the group the same as the product of the numbers of elements each of these generators alone can generate?
If so,is it also true that for two finite groups, if they have the same minimum number of generators needed to construct and that each generator alone can generate the same number of elements corresponding to another group, then the two groups are isomorphic?
For example, if I two groups that need 2 generators,and one of the generator generates 3 elements(including e) and another one generates 6 elements, then (3-1)(6-1)=10 is the number of elements of the group and the two groups are isomorphic to each other?
I just started learning group theory a few days ago, so hope you don't mind if I actually ask something stupid.
Briefly , My attempt is that the set generated by a generator is a cyclic group, and that the direct product of those cyclic groups should be isomorphic to the original group itself.
group-theory
asked Sep 13 at 0:19
Johnny Chen
1026
1
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For a finite group that needs n generator to construct, is the number of elements of the group the same as the product of the numbers of elements each of these generators alone can generate?
If so,is it also true that for two finite groups, if they have the same minimum number of generators needed to construct and that each generator alone can generate the same number of elements corresponding to another group, then the two groups are isomorphic?
For example, if I two groups that need 2 generators,and one of the generator generates 3 elements(including e) and another one generates 6 elements, then (3-1)(6-1)=10 is the number of elements of the group and the two groups are isomorphic to each other?
I just started learning group theory a few days ago, so hope you don't mind if I actually ask something stupid.
Briefly , My attempt is that the set generated by a generator is a cyclic group, and that the direct product of those cyclic groups should be isomorphic to the original group itself.
group-theory
asked Sep 13 at 0:19
Johnny Chen
1026
For a finite group that needs n generator to construct, is the number of elements of the group the same as the product of the numbers of elements each of these generators alone can generate?
If so,is it also true that for two finite groups, if they have the same minimum number of generators needed to construct and that each generator alone can generate the same number of elements corresponding to another group, then the two groups are isomorphic?
For example, if I two groups that need 2 generators,and one of the generator generates 3 elements(including e) and another one generates 6 elements, then (3-1)(6-1)=10 is the number of elements of the group and the two groups are isomorphic to each other?
I just started learning group theory a few days ago, so hope you don't mind if I actually ask something stupid.
Briefly , My attempt is that the set generated by a generator is a cyclic group, and that the direct product of those cyclic groups should be isomorphic to the original group itself.
group-theory
group-theory
asked Sep 13 at 0:19
Johnny Chen
1026
asked Sep 13 at 0:19
Johnny Chen
1026
asked Sep 13 at 0:19
Johnny Chen
1026
asked Sep 13 at 0:19
Johnny Chen
1026
asked Sep 13 at 0:19
Johnny Chen
1026
1026
1
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00
add a comment |Â
1
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00
1
1
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
$S_n$ can be generated by $sigma = (12)$ and $tau = (12dots n)$. The order of $sigma$ is $2$ and the order of $tau$ is $n$. However, the number of elements in $S_n$ is $n!$.
answered Sep 13 at 0:44
matt stokes
57729
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
$S_n$ can be generated by $sigma = (12)$ and $tau = (12dots n)$. The order of $sigma$ is $2$ and the order of $tau$ is $n$. However, the number of elements in $S_n$ is $n!$.
answered Sep 13 at 0:44
matt stokes
57729
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
add a comment |Â
up vote
7
down vote
accepted
$S_n$ can be generated by $sigma = (12)$ and $tau = (12dots n)$. The order of $sigma$ is $2$ and the order of $tau$ is $n$. However, the number of elements in $S_n$ is $n!$.
answered Sep 13 at 0:44
matt stokes
57729
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
$S_n$ can be generated by $sigma = (12)$ and $tau = (12dots n)$. The order of $sigma$ is $2$ and the order of $tau$ is $n$. However, the number of elements in $S_n$ is $n!$.
answered Sep 13 at 0:44
matt stokes
57729
$S_n$ can be generated by $sigma = (12)$ and $tau = (12dots n)$. The order of $sigma$ is $2$ and the order of $tau$ is $n$. However, the number of elements in $S_n$ is $n!$.
answered Sep 13 at 0:44
matt stokes
57729
answered Sep 13 at 0:44
matt stokes
57729
answered Sep 13 at 0:44
matt stokes
57729
answered Sep 13 at 0:44
matt stokes
57729
57729
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
add a comment |Â
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
Thank you, I know what I got wrong now. I think that the product of the order of the generators only tells at least how many elements there are. Maybe, the number elements is the maximum bound of the product of order of distinct generators that work.
– Johnny Chen
Sep 13 at 1:25
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
No problem! This is a good question to ask. The order of the group being an upper bound for the product of generators of course depends on the generating set. For example, if $G$ is a group then the set $G$ generates itself and so the product of the orders of all elements of $G$ is probably much larger than the order of $G$. Given a finite group, can you ever pick a minimal generating set such that the product of the orders of the generators (there has to be an easier way to say that) is larger than the order of the group? I need to think about this some more.
– matt stokes
Sep 13 at 1:56
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
Oh, you are right. How could I not notice that. Let me arrange my argument again. What I had in mind should be that the order of the group is the upper bound for the product of those generators that the intersection of the elements sets each generator generate is empty. The elements each generator generate must be all different from each other.
– Johnny Chen
Sep 13 at 2:11
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
oh yeah， except for the identity. I am too new and to group. Too immature in this area.
– Johnny Chen
Sep 13 at 2:29
add a comment |Â
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1
The situation cannot be as simple as your last paragraph suggests, for that would mean all finite groups are abelian (evidently false).
– hardmath
Sep 13 at 0:31
@hardmath, you are right, I thought something wrong about the direct product. The operation is definitely not as simple.
– Johnny Chen
Sep 13 at 1:00