# Linear algebra of finite abelian groups

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.
(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