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Oct
1
comment Linear algebra of finite abelian groups
(To be more explicit, your answer is no longer valid because the set of generators you produce for $H$ is an irredundant set of generators, but not a basis)
Oct
1
comment Linear algebra of finite abelian groups
(update) yes, {2,3} is both non-redundant and a basis for $\mathbb{Z}/6 \mathbb{Z}$. It is not a basis for $\mathbb{Z}$ since $6 \in (2) \cap (3)$.
Oct
1
answered Constructing a basis for finite abelian groups
Oct
1
awarded  Commentator
Oct
1
comment Linear algebra of finite abelian groups
I have fixed my question with a different notion of "basis" for a finite abelian group. Sorry it took me so long. This is the question I really meant to ask. Does this make more sense to you?
Oct
1
revised Linear algebra of finite abelian groups
added 215 characters in body
Sep
26
awarded  Teacher
Sep
26
accepted Splitting exact sequences of finite abelian groups
Sep
26
answered Splitting exact sequences of finite abelian groups
Sep
21
comment Smallest pure subgroup containing a fixed subgroup
I asked the same question on MO before you gave this argument, and it turns out that the answer is negative. Here is the link with a counterexample mathoverflow.net/questions/107768/…
Sep
21
revised Smallest pure subgroup containing a fixed subgroup
added 1 characters in body
Sep
21
awarded  Editor
Sep
21
revised Smallest pure subgroup containing a fixed subgroup
added 139 characters in body
Sep
21
comment Smallest pure subgroup containing a fixed subgroup
@Jack Schmidt Can you say something more on how you find this $h_1$? Moreover, how do we know that $g_2 \notin \langle h_1 \rangle$ ?
Sep
20
comment Smallest pure subgroup containing a fixed subgroup
@Hagen von Eitzen, the intersection of pure subgroups is not pure, so the title is a bit misleading. What I mean more precisely with "smallest" is explained in the question.
Sep
20
asked Smallest pure subgroup containing a fixed subgroup
Sep
19
awarded  Scholar
Sep
19
comment Splitting exact sequences of finite abelian groups
Thanks! Yes, it seems to me that Chinese tells that GCD is equivalent to the existence of $k_i$ satisfying (1) and (2).
Sep
18
comment Splitting exact sequences of finite abelian groups
Thanks for your answer. Sorry but this looks to me just as the definition. Is it obvious that any of the three equivalent conditions proves (or disproves) the statement in my question?
Sep
18
awarded  Student