# How to show that the kernel of a map $f:\mathbb{Z}^m\to G$ is free of rank $m$ when $G$ is finite?

In my studies for my quals next week, I am trying to prove the following problem:

Let $G$ be a finite Abelian group. Let $f:\mathbb{Z}^m\to G$ be a surjection of Abelian groups. We may think of $f$ as a homomorphism of $\mathbb{Z}$ modules. Let $K$ be the kernel of $f$.
(a) Prove that $K$ is isomorphic to $\mathbb{Z}^m$.
(b) We can therefore write the inclusion map $K\to \mathbb{Z}^m$ as $\mathbb{Z}^m\to \mathbb{Z}^m$ and represent it by an $m\times m$ matrix $A$. Prove that $|\det A| = |G|$.

I can prove part (b) using Smith-Normal form, but I can't seem to do part (a). It seems true because you need to 'eliminate' all parts of the free part to get to a finite group, but I can't seem to justify it. Is there some sort of 'change of basis' I could do to make it obvious, as in part (b), or is some sort of standard manipulation necessary that I'm not seeing.

Thanks so much for your time!

• Do you know the fundamental theorem on abelian groups? – Bernard Sep 4 '17 at 17:41
• @Bernard Indeed. So, I could write $G\simeq \bigoplus_i \mathbb{Z}/(p_i^{k_i})$. I suppose the guess would be to say that the kernel is generated freely by the inverse image of each of the generators, but I couldn't see why that was true. – Laarz Sep 4 '17 at 17:44
• The theorem says more than that: for a finite group: there exist a basis $e_1, \dots, e_m$ of $\mathbf Z^m\;$ and positive integers $a_1,\dots, a_m$ such that: $1)$ for each $p=1,\dots m-1, \;a_i\mid a_{i+1}$, and $\;2)$, $\{a_1e_1,\dots, a_me_m$ is a basis of $K$. So the matrix in this basis is rather easy to write, and its determinant still easier to compute. B.t.w. the coefficients are the invariant factors of $G$ (what you wrote are the elementary divisors). – Bernard Sep 4 '17 at 17:55
• I’m really rusty on this stuff. Couldn’t you just tensor with $\Bbb Q$? – Lubin Sep 4 '17 at 22:36
• O_O. Amazing. Thanks, Lubin~ – Laarz Sep 5 '17 at 16:26

$\require{begingroup}\begingroup\DeclareMathOperator{\rnk}{rank}$If $n = |G|$ then $n \mathbf{Z}^m \subseteq \ker f \subseteq \mathbf{Z}^m$. This shows that $$\rnk(n \mathbf{Z}^m) = m \le \rnk(\ker f) \le \rnk(\mathbf{Z}^m) = m.$$ So $\rnk(\ker f) = m$. Now since $\ker f$ is a subgroup of a finitely generated free abelian group, we have a lemma that says that $\ker f$ is free.
We can also use the theorem that says that if $\ker f \subseteq \mathbf{Z}^m$ and $\rnk(\ker f) = r$ then there should exist a basis $x_1,\dots,x_m$ for $\mathbf{Z}^m$ and positive integers $a_1 \mid a_2 \mid \dots \mid a_r$ such that $a_1x_1,\dots,a_rx_r$ is a basis for $\ker f$. Since $G$ is finite, this automatically implies $r = m$ and we can also infer part b) since $|\det A| = a_1\cdots a_m = |G|$. $\endgroup$
• I had thought about this, even noting that each $e_i = (0,\ldots, 1, \ldots, 0)$, where the $1$ is in the $i$th component, would have some order, say $n_i$, which would then show $\bigoplus_i n_i\mathbb{Z}\subseteq K \subseteq \mathbb{Z}^m$, but I couldn't get anywhere with this. – Laarz Sep 4 '17 at 17:41
• @Laarz This shows that $\operatorname{rank}(K) = m$. You then have to use the lemma that a subgroup of a finitely generated free abelian group is also free. I don't see how to avoid using the full strength of that lemma (i.e. whether or not having $K$ come from a map $\mathbf{Z}^m \to G$ gives an easier proof than just having $K$ be any old subgroup). – Trevor Gunn Sep 4 '17 at 17:45