Let \(G\) be a finite abelian group. Let \(f:\mathbb Z^m\rightarrow G\) be a surjection of abelian groups. We may think of \(f\) as a homomorphism of \(\mathbb Z\)-modules. Let \(K\) be the kernel of \(f\).
Prove that \(K\) is isomorphic to \(\mathbb Z^m\).
We can therefore write the inclusion map \(K\rightarrow\mathbb Z^m\) as \(\mathbb Z^m\rightarrow\mathbb Z^m\) and represent it by an \(m\times m\) integer matrix \(A\). Prove that \(|\det A|=|G|\).
<1>4. After choosing a basis of \(K\), let \[ A:\mathbb Z^m\longrightarrow\mathbb Z^m \] be the integer matrix representing the inclusion \(K\hookrightarrow\mathbb Z^m\). Then \[ \mathbb Z^m/A\mathbb Z^m\cong G. \]
<1>5. Let \[ UAV=\operatorname{diag}(d_1,\dots,d_m) \] be a Smith normal form of \(A\), with \(U,V\in\operatorname{GL}_m(\mathbb Z)\) and each \(d_i>0\). Then \[ |G|=d_1\cdots d_m. \]
<1>6. Finally, \[ |\det A|=|G|. \]
Solution
1. The subgroup \(K\le \mathbb Z^m\) is a free abelian group of rank at most \(m\).
Every subgroup of a finitely generated free abelian group is free abelian of rank at most the rank of the ambient group. Thus \[ K\cong \mathbb Z^r \] for some \(0\le r\le m\).
2. In fact \(r=m\).
The first isomorphism theorem gives \[ \mathbb Z^m/K\cong G. \] Since \(G\) is finite, tensoring the short exact sequence \[ 0\longrightarrow K\longrightarrow \mathbb Z^m\longrightarrow G\longrightarrow0 \] with \(\mathbb Q\) gives \[ K\otimes_{\mathbb Z}\mathbb Q\cong \mathbb Q^m, \] because \(G\otimes_{\mathbb Z}\mathbb Q=0\). If \(K\cong\mathbb Z^r\), then \[ K\otimes_{\mathbb Z}\mathbb Q\cong\mathbb Q^r. \] Therefore \(r=m\).
3. Hence\[ K\cong\mathbb Z^m. \]
Under the chosen identification \(K\cong\mathbb Z^m\), the image of \(A\) is precisely the subgroup \(K\). Hence \[ \mathbb Z^m/A\mathbb Z^m \cong \mathbb Z^m/K \cong G. \]
The quotient in 4 is finite, so \(A\) has full rank and all Smith invariants \(d_i\) are nonzero. Multiplication by the unimodular matrices \(U\) and \(V\) does not change the isomorphism type of the cokernel, so \[ G\cong \mathbb Z^m/A\mathbb Z^m \cong \bigoplus_{i=1}^m \mathbb Z/d_i\mathbb Z. \] Therefore \[ |G|=\prod_{i=1}^m d_i. \]
Since \(U\) and \(V\) are unimodular, \[ |\det U|=|\det V|=1. \] Thus \[ |\det A| =|\det(UAV)| =\prod_{i=1}^m d_i =|G| \] by 5.