Number of generators of subgroup Announcing the arrival of Valued Associate #679: Cesar...
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Number of generators of subgroup
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Torsion subgroupOn the minimal number of generators of a finite groupBound number of generators of a subgroup of a nilpotent group?Minimal number of generators for a finitely generated abelian $p$-groupA question on finitely generated Abelian groups with a minimal number of generatorsFactoring an Abelian groupThe number of internal direct summands of a finitely generated abelian groupFree group generated by two generators is isomorphic to product of two infinite cyclic groupsAlternative proof of the Fundamental Theorem of Abelian Groups??Hungerford Chapter 2 Section 2 Problem 2 WITHOUT using the structure theorem of finite abelian groups
$begingroup$
I am trying to prove the following.
let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?
Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbb{Z}$ and $H=2mathbb{Z}$.
I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.
group-theory abelian-groups
asked 5 hours ago
CharlesCharles
582420
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following.
let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?
Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbb{Z}$ and $H=2mathbb{Z}$.
I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.
group-theory abelian-groups
asked 5 hours ago
CharlesCharles
582420
$endgroup$
$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
2
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago
add a comment |
$begingroup$
I am trying to prove the following.
let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?
Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbb{Z}$ and $H=2mathbb{Z}$.
I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.
group-theory abelian-groups
asked 5 hours ago
CharlesCharles
582420
$endgroup$
I am trying to prove the following.
let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?
Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbb{Z}$ and $H=2mathbb{Z}$.
I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.
group-theory abelian-groups
group-theory abelian-groups
asked 5 hours ago
CharlesCharles
582420
asked 5 hours ago
CharlesCharles
582420
edited 5 hours ago
Charles
asked 5 hours ago
CharlesCharles
582420
asked 5 hours ago
CharlesCharles
582420
asked 5 hours ago
CharlesCharles
582420
582420
$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
2
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago
add a comment |
$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
2
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago
$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
2
2
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$
answered 5 hours ago
MelodyMelody
1,42212
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$
answered 5 hours ago
MelodyMelody
1,42212
$endgroup$
add a comment |
$begingroup$
No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$
answered 5 hours ago
MelodyMelody
1,42212
$endgroup$
add a comment |
$begingroup$
No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$
answered 5 hours ago
MelodyMelody
1,42212
$endgroup$
No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$
answered 5 hours ago
MelodyMelody
1,42212
answered 5 hours ago
MelodyMelody
1,42212
answered 5 hours ago
MelodyMelody
1,42212
answered 5 hours ago
MelodyMelody
1,42212
1,42212
add a comment |
add a comment |
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$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago
2
$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago
$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago