Computation of a homology group of a simple complex $0\rightarrow \mathbb{Z}^l \rightarrow \mathbb{Z}^n \rightarrow \mathbb{Z}^m\rightarrow 0$

Consider the following sequence of abelian groups, where $f\circ g = 0$.

$$0\longrightarrow \mathbb{Z}^l \overset{g}{\longrightarrow}\mathbb{Z}^n\overset{f}{\longrightarrow}\mathbb{Z}^m\longrightarrow 0 \tag{1}$$

Suppose $f$ and $g$ are explicitly given by matrices. I would like to determine the structure of $H = Ker(f)/Im(g)$. By the fundamental theorem of abelian groups, it is a direct of sum of cyclic groups: $H = \sum C_i$. How can we compute the order of each $C_i$?

• Dear Makoto, The answer depends on the maps $f$ and $g$. Are you asking about the procedure for making the computation? Regards, – Matt E Dec 20 '13 at 1:52
• @MattE Dear Matt, Yes, I would like to know the algorithm for the computation. – Makoto Kato Dec 20 '13 at 1:57

So you want to find the image and the kernel of matrices over $\mathbb Z$. You should instinctively want to use what you already know about linear algebra, but you're not over a field. It turns out that the kernel is free and is precisely the null space if you consider the matrix over $\mathbb R$. However, for image you aren't so lucky. However, if you put the matrix $g$ in Smith-Normal form, it should happen that the change of basis you've done is over integers and it allows you to easily read off the image.
• [It turns out that the kernel is free and is precisely the null space if you consider the matrix over $\mathbb{R}$] Could you explain how we can find a free $\mathbb{Z}$-basis of the kernel of $f$? Regards, – Makoto Kato Dec 21 '13 at 1:51