Is this true for about abelian generating sets?












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    What exactly is your definition of "basis" here?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:11






  • 2




    $begingroup$
    Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
    $endgroup$
    – Yanko
    Jan 4 at 16:21










  • $begingroup$
    Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:32






  • 2




    $begingroup$
    Your definition of a basis is the same as a generating set...
    $endgroup$
    – verret
    Jan 4 at 18:38










  • $begingroup$
    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.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 21:49


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    What exactly is your definition of "basis" here?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:11






  • 2




    $begingroup$
    Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
    $endgroup$
    – Yanko
    Jan 4 at 16:21










  • $begingroup$
    Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:32






  • 2




    $begingroup$
    Your definition of a basis is the same as a generating set...
    $endgroup$
    – verret
    Jan 4 at 18:38










  • $begingroup$
    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.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 21:49
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 18:14









Shaun

10.3k113686




10.3k113686










asked Jan 4 at 16:02









I_wil_break_wallI_wil_break_wall

855




855








  • 3




    $begingroup$
    What exactly is your definition of "basis" here?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:11






  • 2




    $begingroup$
    Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
    $endgroup$
    – Yanko
    Jan 4 at 16:21










  • $begingroup$
    Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:32






  • 2




    $begingroup$
    Your definition of a basis is the same as a generating set...
    $endgroup$
    – verret
    Jan 4 at 18:38










  • $begingroup$
    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.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 21:49
















  • 3




    $begingroup$
    What exactly is your definition of "basis" here?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:11






  • 2




    $begingroup$
    Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
    $endgroup$
    – Yanko
    Jan 4 at 16:21










  • $begingroup$
    Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
    $endgroup$
    – Torsten Schoeneberg
    Jan 4 at 16:32






  • 2




    $begingroup$
    Your definition of a basis is the same as a generating set...
    $endgroup$
    – verret
    Jan 4 at 18:38










  • $begingroup$
    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.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 21:49










3




3




$begingroup$
What exactly is your definition of "basis" here?
$endgroup$
– Torsten Schoeneberg
Jan 4 at 16:11




$begingroup$
What exactly is your definition of "basis" here?
$endgroup$
– Torsten Schoeneberg
Jan 4 at 16:11




2




2




$begingroup$
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
$endgroup$
– Yanko
Jan 4 at 16:21




$begingroup$
Perhaps by a basis you mean a minimal generating set? If that's so it's not hard to show that such exists.
$endgroup$
– Yanko
Jan 4 at 16:21












$begingroup$
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
$endgroup$
– Torsten Schoeneberg
Jan 4 at 16:32




$begingroup$
Just to check: According to your definition, in your example, $lbrace 1rbrace$ is also a basis of $Bbb Z_6$, correct?
$endgroup$
– Torsten Schoeneberg
Jan 4 at 16:32




2




2




$begingroup$
Your definition of a basis is the same as a generating set...
$endgroup$
– verret
Jan 4 at 18:38




$begingroup$
Your definition of a basis is the same as a generating set...
$endgroup$
– verret
Jan 4 at 18:38












$begingroup$
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.
$endgroup$
– Pierre-Guy Plamondon
Jan 4 at 21:49






$begingroup$
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.
$endgroup$
– Pierre-Guy Plamondon
Jan 4 at 21:49












1 Answer
1






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4












$begingroup$

No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 22:00














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1 Answer
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4












$begingroup$

No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 22:00


















4












$begingroup$

No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 22:00
















4












4








4





$begingroup$

No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.






share|cite|improve this answer









$endgroup$



No: in $mathbb{Z}_{12}$, $S={2,3}$ generates $mathbb{Z}_{12}$, but ${2}$ and ${3}$ don't, and $mathbb{Z}_{12} neq langle 2 rangle times langle 3 rangle$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 16:47









Pierre-Guy PlamondonPierre-Guy Plamondon

8,90511739




8,90511739












  • $begingroup$
    This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 22:00




















  • $begingroup$
    This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
    $endgroup$
    – Pierre-Guy Plamondon
    Jan 4 at 22:00


















$begingroup$
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
$endgroup$
– Pierre-Guy Plamondon
Jan 4 at 22:00






$begingroup$
This answer is correct only if one uses the definition of a basis appearing in the original version of the question.
$endgroup$
– Pierre-Guy Plamondon
Jan 4 at 22:00




















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