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Let $\phi:G \to H$ be a surjective homomorphism of finite abelian groups, and

let $g_1, \ldots, g_n$ be an irredundant set of generators (from now on, a basis) for $G$. be a basis for $G$, meaning a set of elements with the property that $\langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle=G$ (note they have to be in direct sum).

To any basis $h_1, \ldots, h_k$ for $H$, we can associate a matrix $a$ with entries in $\mathbb{Z}$ representing $\phi$, defined by $\phi(g_i)= \sum_{j=1}^k a_{i j} h_j$.

I would like to prove the following statement. Up to permuting the $g_i$'s, we can find a basis for $H$ (see above) such that the left lower triangle of the associated matrix $a$ is made of ones on the diagonal and zeroes elsewhere. (In the previous sentence, I am assuming that the length of the basis for $H$ must satisfy $k \leq n$). More precisely, such that $a_{n-k+ i, i}=1$ and $a_{n-k+j,i}=0$ for $j>i$.

Do you know of a good reference for this kind of elementary problems? Namely, linear algebra among finite, or finitely generated, abelian groups?

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Do you consider $\{2,3\}$ to be a non-redundant set of generators of $(\mathbf Z/6\mathbf Z,{+})$? – Marc van Leeuwen Sep 7 '12 at 10:10
And what if the images of all the $g_i$ are the same? – Marc van Leeuwen Sep 7 '12 at 10:12
Yes, for me irredundant means minimal, not necessarily of minimal cardinality. – calc Sep 7 '12 at 10:13
Do you mean in the case when $H$ is cyclic? In that case the statement seems easy to me. – calc Sep 7 '12 at 10:16
@calc: does that mean you consider $\{2, 3\}$ to be a non-redundant set of generators for $(\mathbf{Z}, +)$? – Hurkyl Sep 7 '12 at 10:35

If I understand your question (which seems to be the most difficult part) the following simple procedure provides a positive answer, without requiring any group theory. Take the sequence of $g_i$ and while possible eject any element whose image can be expressed in terms of the images of the other (non-ejected) elements. When this terminates, you're left with a sequence of elements whose images generate $H$, and such that removing any one of them would destroy this property. But then these images are by definition a non-redundant sequence of generators of $H$, and can be chosen as the $h_i$. On the $G$ side, tack all ejected elements to the end of the sequence of non-ejected ones. Note that this makes the initial part of the matrix equal to an identity matrix, better than what you asked for.

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You are right. My question was silly. Thanks and sorry. – calc Sep 7 '12 at 11:53
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? – calc Oct 1 '12 at 7:24
(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) – calc Oct 1 '12 at 12:09

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