If $f: V \to W$ is a surjective homomorphism of vector spaces, and we have fixed a basis for $V$, it is always possible to find a basis for $W$ such that the matrix associated to $\phi$ in the two bases is triangular with ones on the "diagonal" (what I mean with this is explained more precisely later, in the case of abelian groups), up to possibly permuting the chosen basis of $V$. This motivates the following question, in the realm of finite abelian groups.

Let $H$ be a finite abelian group with a fixed basis $h_1, \ldots, h_n$, with "basis" here we mean elements that satisfy the property $\langle h_1 \rangle \oplus \ldots \oplus \langle h_n \rangle=H$. Let $\phi:H \to G$ be a homomorphism of finite abelian groups. Let us assume that $\phi$ is surjective, so $\phi(h_1), \ldots, \phi(h_n)$ generate $G$.

Is it possible to find a basis for $G$, and a permutation of the elements $h_1, \ldots, h_n$, such that the matrix associated with $\phi$ in the two given basis has a triangular form with $1$ on the diagonal?

More precisely, can we find a basis $g_1, \ldots, g_m$ for $G$, with $m \leq n$, and a permutation $\sigma \in \mathfrak{S}_n$ such that we can write

$\phi(h_{\sigma(i)}) = a_{i,1} g_1 + \ldots+ a_{i,m} g_m$

with $a_{i,j}=0$ when $j>i$, and $a_{i,i}=1$?

(I asked a similar question on math.stackexchange, but I spelled it differently and in a silly/wrong way so that the answer in that case was trivial)

  • 1
    $\begingroup$ Why don't you edit your question in math.stackexchange so that it says what you really mean? $\endgroup$ – Fernando Muro Sep 30 '12 at 20:34
  • $\begingroup$ Here is my edited question in math.stackexchange. math.stackexchange.com/questions/192322/… $\endgroup$ – calc Oct 1 '12 at 7:26
  • $\begingroup$ What du you mean with a triangular matrix or a diagonal if V and W have different dimensions? $\endgroup$ – Matthias Ludewig Oct 3 '12 at 11:36
  • 1
    $\begingroup$ I believe the answer to your question is yes, which should be possible to work out from the structure theorem for finite abelian groups. $\endgroup$ – Matthias Ludewig Oct 3 '12 at 11:43
  • $\begingroup$ @Kofi I mean the same that I explain later in the case of abelian groups $\endgroup$ – calc Oct 3 '12 at 12:09

I think I can prove this in the case when $G$ is a $p$-group. Assume $\phi(h_1), \ldots, \phi(h_m)$ are irredundant generators: if any of them is removed they are no longer a generating set for $G$. There must be an element of maximal order, and assume after possibly permuting that it is $\phi(h_1)$. Then the cyclic group generated by $\phi(h_1)$ is a direct summand of $G$, thus we can take $\phi(h_1)$ as the first element of the basis for $G$ and pass to the quotient $G/\langle \phi_(h_1) \rangle$. Consider now the projection of $\phi(h_2), \ldots, \phi(h_m)$, and suppose $\phi(h_2)$ has maximal order in the quotient (thus again it is a direct summand in the quotient): then take $\phi(h_2)$ as second element element of the basis for $G$, and so on.

To my big surprise, I think this property need not be true when $G$ is not a $p$-group, although I am not sure I have a complete proof for this. My example is in the case $G=\mathbb{Z}_2 \oplus \mathbb{Z}_8$ $\oplus \mathbb{Z}_3 \oplus \mathbb{Z}27$ with the elements $\phi(h_1)= (1,2,0,1)$ and $\phi(h_2)=(0,1,1,3)$.

| cite | improve this answer | |
  • 2
    $\begingroup$ It is quite strange to answer your own question... $\endgroup$ – Jérémy Blanc Oct 3 '12 at 13:19
  • $\begingroup$ You are right, but it seems nobody else cares. I have the feeling that everyone is convinced it is trivially true (which may very well be the case). $\endgroup$ – calc Oct 3 '12 at 13:24
  • $\begingroup$ It is a shame people can not vote for their own question and answer :) $\endgroup$ – calc Oct 3 '12 at 13:32
  • $\begingroup$ Can't you just use the Chinese Remainder Theorem to derive the general result from the result for $p$-groups? $\endgroup$ – RP_ Oct 3 '12 at 13:50
  • 4
    $\begingroup$ It is not strange to answer your own question: It is actually fairly common. $\endgroup$ – Steven Gubkin Oct 17 '12 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.