Is this true for about abelian generating sets?
Given a finite abelian group $G$ which is genertaed by a set $S$.
Question : Is there always exists a $S' subseteq S$ which is a basis of $G$?
Seems true to me for example $mathbb{Z}_6$, take $S = {1,2,3}$ contains a subset ${2,3}$ which is a basis.
A set $B$ is said to be a basis of a abelian group $G$ if $G=langle B rangle$.
group-theory finite-groups abelian-groups
edited Jan 4 at 18:14
Shaun
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
add a comment |
Given a finite abelian group $G$ which is genertaed by a set $S$.
Question : Is there always exists a $S' subseteq S$ which is a basis of $G$?
Seems true to me for example $mathbb{Z}_6$, take $S = {1,2,3}$ contains a subset ${2,3}$ which is a basis.
A set $B$ is said to be a basis of a abelian group $G$ if $G=langle B rangle$.
group-theory finite-groups abelian-groups
edited Jan 4 at 18:14
Shaun
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
3
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
2
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
2
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49
add a comment |
Given a finite abelian group $G$ which is genertaed by a set $S$.
Question : Is there always exists a $S' subseteq S$ which is a basis of $G$?
Seems true to me for example $mathbb{Z}_6$, take $S = {1,2,3}$ contains a subset ${2,3}$ which is a basis.
A set $B$ is said to be a basis of a abelian group $G$ if $G=langle B rangle$.
group-theory finite-groups abelian-groups
edited Jan 4 at 18:14
Shaun
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
Given a finite abelian group $G$ which is genertaed by a set $S$.
Question : Is there always exists a $S' subseteq S$ which is a basis of $G$?
Seems true to me for example $mathbb{Z}_6$, take $S = {1,2,3}$ contains a subset ${2,3}$ which is a basis.
A set $B$ is said to be a basis of a abelian group $G$ if $G=langle B rangle$.
group-theory finite-groups abelian-groups
group-theory finite-groups abelian-groups
edited Jan 4 at 18:14
Shaun
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
edited Jan 4 at 18:14
Shaun
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
edited Jan 4 at 18:14
Shaun
8,820113681
edited Jan 4 at 18:14
Shaun
8,820113681
edited Jan 4 at 18:14
Shaun
8,820113681
8,820113681
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
asked Jan 4 at 16:02
I_wil_break_wallI_wil_break_wall
284
284
3
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
2
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
2
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49
add a comment |
3
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
2
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
2
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49
3
3
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
2
2
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
2
2
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49
add a comment |
1 Answer
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No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
add a comment |
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1 Answer
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No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
add a comment |
No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
add a comment |
No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
answered Jan 4 at 16:47
Pierre-Guy PlamondonPierre-Guy Plamondon
8,76011639
8,76011639
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
add a comment |
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
– Pierre-Guy Plamondon
Jan 4 at 22:00
add a comment |
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3
What exactly is your definition of "basis" here?
– Torsten Schoeneberg
Jan 4 at 16:11
2
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
– Yanko
Jan 4 at 16:21
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
– Torsten Schoeneberg
Jan 4 at 16:32
2
Your definition of a basis is the same as a generating set...
– verret
Jan 4 at 18:38
Your new definition of a basis is not equivalent to your old one. As verret writes, the new one is the same as the definition of a generating set.
– Pierre-Guy Plamondon
Jan 4 at 21:49