The number of elements of a finite group which is a quotient of a finitely generated free abelian group

Question

Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\alpha_1,\cdots,\alpha_m$ be a finite sequence of elements of $G$. Suppose $\alpha_i = \sum_j a_{ij} \omega_j$ for $i = 1,\cdots, m$. Let $H$ be the subgroup of $G$ generated by $\alpha_1,\cdots,\alpha_m$. Can we compute $|G/H|$ by the data $a_{ij}, 1 \le i \le m, 1 \le n\le n$? If yes, how?

Answer 1

Yes:

1) if $\;m<n\;$ then the index is "obviously" (why?) infinite ;

2) If $\;m=n\;$ then form the square $\;n\times n\;$ matrix $\;A=(a_{ij})\;$ , then

$$|G/H|=[G:H]=\left|\det A\right|$$

The proof of (b) is a rather beautiful subject within "finitely generated abelian groups", which include the canonical Smith form for (integer) matrices and stuff. You can google this to read more about it.

Answer 2

Let $x_1,\cdots, x_m$ be a sequence of elements of an abelian group $G$. We denote by $[x_1,\cdots,x_m]$ the subgroup of $G$ generated by $x_1,\cdots, x_m$.

Lemma 1 Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\alpha_1,\cdots,\alpha_m$ be a finite sequence of elements of $G$. Then there exists an algorithm to find $\beta_1,\cdots,\beta_n \in G$ such that $[\alpha_1,\cdots,\alpha_m] = [\beta_1,\cdots,\beta_n]$ and $\beta_i \in [\omega_1,\cdots,\omega_i]$ for all $i = 1,\cdots,n$. Moreover we may choose each $\beta_i$ such that its coefficient with respect to $\omega_i$ is non-negative.

Proof: We use induction on the rank $n$ of $G$. Let $\alpha_i = \sum_j a_{ij}\omega_j$ for $i = 1,\cdots,m$. Suppose $n = 1$. Then $\alpha_1 = a_{11}\omega_1,\cdots, \alpha_m = a_{m1}\omega_1$. If all of $a_{11},\cdots,a_{m1}$ are zero, let $\beta_1 = 0$ and we are done. Otherwise let $d =$ gcd$(a_{11},\cdots,a_{m1})$. Let $\beta_1 = d\omega_1$ and we are done.

Suppose $n \gt 1$. If all of $a_{1n},\cdots,a_{mn}$ are zero, let $\beta_n = 0$. Since $[\alpha_1, \cdots\alpha_m] \subset [\omega_1,\cdots,\omega_{n-1}]$, by the induction hypothesis, there exists an algorithm to find $\beta_1,\cdots,\beta_{n-1} \in [\omega_1,\cdots,\omega_{n-1}]$ such that $[\alpha_1,\cdots,\alpha_m] = [\beta_1,\cdots,\beta_{n-1}]$ and $\beta_i \in [\omega_1,\cdots,\omega_i]$ for all $i = 1,\cdots,{n-1}$. Hence $\beta_1,\cdots,\beta_n$ satisfies the condition of the lemma. Suppose not all of $a_{1n},\cdots,a_{mn}$ are zero. Let $d =$ gcd$(a_{1n},\cdots,a_{mn})$. Let $a_{in} = b_id$ for $i = 1,\cdots,m$. There exist integers $c_1,\cdots,c_m$ such that $d = c_1a_{1n} + \cdots + c_ma_{mn}$. Let $\beta_n = c_1\alpha_1 + \cdots + c_m\alpha_m$. Let $\gamma_i = \alpha_i- b_i\beta_n$ for $i = 1,\cdots, m$. Then that $[\gamma_1,\cdots,\gamma_m, \beta_n] = [\alpha_1,\cdots,\alpha_m]$. Since $\gamma_i \in [\omega_1,\cdots,\omega_{n-1}]$ for $i = 1,\cdots, m$, by the induction hypothesis, there exists an algorithm to find $\beta_1,\cdots,\beta_{n-1} \in [\omega_1,\cdots,\omega_{n-1}]$ such that $[\gamma_1,\cdots,\gamma_m] = [\beta_1,\cdots,\beta_{n-1}]$ and $\beta_i \in [\omega_1,\cdots,\omega_i]$ for all $i = 1,\cdots,n-1$. Hence $\beta_1,\cdots,\beta_n$ satisfies the condition of the lemma. QED

Lemma 2 Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\alpha_1,\cdots,\alpha_n \in G$ be such that

$\alpha_1 = a_{11}\omega_1$

$\alpha_2 = a_{21}\omega_1 + a_{22}\omega_2$

. . .

$\alpha_i = a_{i1}\omega_1 + \cdots + a_{ii}\omega_i$

. . .

$\alpha_n = a_{n1}\omega_1 + \cdots + a_{nn}\omega_n$

Suppose $a_{ii} \ne 0$ for all $i = 1,\cdots, n$. Let $H = [\alpha_1,\cdots,\alpha_n]$. Let $x = x_1\omega_1 + \cdots + x_i\omega_i \in H$ for some $i$ such that $1 \le i \le n$, where $x_1,\cdots,x_i$ are integers. Then $x_i$ is divisible by $a_{ii}$.

Proof: There exist integers $b_1,\cdots,b_n$ such that $x = b_1\alpha_1 + \cdots + b_n\alpha_n$. Then $n$-th coefficient of $x$ is $b_na_{nn}$ which must be $0$. Hence $b_n = 0$. Similarly we get $b_{i+1} = \cdots = b_n = 0$. Hence $x_i = b_ia_{ii}$. QED

Lemma 3 Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\alpha_1,\cdots,\alpha_n \in G$ be such that

$\alpha_1 = a_{11}\omega_1$

$\alpha_2 = a_{21}\omega_1 + a_{22}\omega_2$

. . .

$\alpha_i = a_{i1}\omega_1 + \cdots + a_{ii}\omega_i$

. . .

$\alpha_n = a_{n1}\omega_1 + \cdots + a_{nn}\omega_n$

Suppose $a_{ii} \gt 0$ for all $i = 1,\cdots, n$. Let $H = [\alpha_1,\cdots,\alpha_n]$. Then $|G/H| = a_{11}\cdots a_{nn}$.

Proof: It suffices to prove that $\{\sum_i r_i\omega_i\ |\ 0 \le r_i \lt a_{ii}, i = 1\cdots,n\}$ is a complete set of representatives for $G/H$.

Let $x = x_1\omega_1 + \cdots + x_n\omega_n \in G$. there exist integers $q_n, r_n$ such that $x_n = a_{nn}q_n + r_n, 0 \le r_n \lt a_{nn}$. Then $x - q_n\alpha_n = y_1\omega_1 +\cdots + y_{n-1}\omega_{n-1} + r_n\omega_n$, where $y_1,\cdots, y_{n-1}$ are integers. Applying this process to $x - q_n\alpha_n$, etc. we get $x - \sum_i q_i\alpha_i = \sum_i r_i\omega_i$, where $0 \le r_i \lt a_{ii}$ for all $i$. Hence $x \equiv \sum_i r_i\omega_i$ (mod $H$).

Let $\sum_i r_i\omega_i \equiv \sum_i s_i\omega_i$ (mod $H$), where $0 \le r_i, s_i \lt a_{ii}$ for all $i$. We need to prove that $r_i = s_i$ for all $i$. By Lemma 2, $r_n = s_n$. Hence $r_1\omega_1 + \cdots + r_{n-1}\omega_{n-1} \equiv s_1\omega_1 + \cdots + s_{n-1}\omega_{n-1}$ (mod $H$). Hence agian by Lemma 2, $r_{n-1} = s_{n-1}$. Repeating this process, we get $r_i = s_i$ for all $i$. QED

Lemma 4 Let $G$ be a finitely generated abelian group. Then $G\otimes_{\mathbb{Z}} \mathbb{Q}= 0$ if and only if $G$ is a finite group.

Proof: Suppose $G$ is a finite group. Let $n$ be the order of $G$. Consider an element $x \otimes r$ of $G\otimes_{\mathbb{Z}} \mathbb{Q}$. Then $x \otimes r = x \otimes n(r/n) = nx\otimes r/n = 0$. Hence $G\otimes_{\mathbb{Z}} \mathbb{Q}= 0$.

Conversely suppose $G\otimes_{\mathbb{Z}} \mathbb{Q}= 0$. Since $G$ is finitely generated, $G$ is a finite product of cyclic groups. Let $G = G_1 \oplus \cdots \oplus G_m$, where each $G_i$ is a cyclic group. Since $G\otimes_{\mathbb{Z}} \mathbb{Q}$ is isomorphic to $G_1\otimes_{\mathbb{Z}} \mathbb{Q} \oplus \cdots \oplus G_m\otimes_{\mathbb{Z}} \mathbb{Q}$, $G_i\otimes_{\mathbb{Z}} \mathbb{Q} = 0$ for all $i$. If $G_i$ is an infinite cyclic group, then $G_i\otimes_{\mathbb{Z}} \mathbb{Q}$ is isomorphic to $\mathbb{Q}$. This is a contradiction. Hence each $G_i$ is a finite cyclic group. Therefore $G$ is a finite group. QED

Lemma 5 Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\alpha_1,\cdots,\alpha_n \in G$ be such that $\alpha_i = \sum_i a_{ij}\omega_j$ for $i = 1,\cdots,n$. Let $H = [\alpha_1,\cdots,\alpha_n]$. Then the following assertions are equivalent.

1. det $(a_{ij}) \ne 0$.

2. $\alpha_1,\cdots,\alpha_n$ is a free basis of $H$.

3. $|G/H|$ is finite.

Proof: $1. \Rightarrow 2.$ Let $A$ be the matrix $(a_{ij})$. Suppose $\alpha_1,\cdots,\alpha_n$ are linearly dependent over $\mathbb{Z}$. Then the row vectors of $A$ are linearly dependent over $\mathbb{Z}$. Hence they are linearly dependent over $\mathbb{Q}$. Hence det $(a_{ij}) = 0$.

$2. \Rightarrow 3.$ Consider the following exact sequence.

$0 \rightarrow H \rightarrow G \rightarrow G/H \rightarrow 0$.

This yields the following exact sequence.

$H\otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow G \otimes_{\mathbb{Z}} \mathbb{Q}\rightarrow (G/H) \otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow 0$.

Since the rank of $H$ is $n$, the map $H\otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow G \otimes_{\mathbb{Z}} \mathbb{Q}$ is surjective. Hence $(G/H) \otimes_{\mathbb{Z}} \mathbb{Q} = 0$. Hence $G/H$ is a finite group by Lemma 4.

$3. \Rightarrow 1.$ Suppose det $(a_{ij}) = 0$. Then the map $H\otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow G \otimes_{\mathbb{Z}}\mathbb{Q}$ in the above sequence is not surjective. Hence $(G/H) \otimes_{\mathbb{Z}} \mathbb{Q} \ne 0$. Hence $G/H$ is an infinite group by Lemma 4. QED

Combining Lemma 1, 2, 5, we get an answer to the question. The following proposition is also useful to compute $G/H$.

Proposition Let $G$ be a finitely generated free abelian group. Let $\omega_1,\cdots,\omega_n$ be its basis. Let $\beta_1,\cdots,\beta_n \in G$ be such that $\beta_i = \sum_i b_{ij}\omega_j$ for $i = 1,\cdots,n$. Let $H = [\beta_1,\cdots,\beta_n]$. If det $(b_{ij}) = 0$, then $|G/H|$ is infinite. If det $(b_{ij}) \ne 0$, then $|G/H| = |$det $(b_{ij})|$.

Proof: If det $(b_{ij}) = 0$, then, by Lemma 4, $|G/H|$ is infinite. Suppose det $(b_{ij}) \ne 0$. By Lemma 5, $\{\beta_1,\cdots,\beta_n\}$ is a free basis of $H$ and $G/H$ is a finite group.

Let $B = (b_{ij})$. Let $\beta$ be the column vector $(\beta_1,\cdots,\beta_n)^t$. Let $\omega$ be the column vector $(\omega_1,\cdots,\omega_n)^t$. Then $\beta = B\omega$.

By Lemma 1, there exist $\alpha_1,\cdots,\alpha_n \in H$ such that $\alpha_i \in [\omega_1,\cdots,\omega_i]$ for all $i$ and $H = [\alpha_1,\cdots,\alpha_n]$. Let $\alpha_i = \sum_j a_{ij}\omega_j$ for all $i$. We may suppose $a_{ii} \ge 0$ for all $i$. Since $G/H$ is a finite group, det $(a_{ij}) \ne 0$ by Lemma 5. Hence $a_{ii} \gt 0$ for all $i = 1,\cdots,n$. By Lemma 2, $|G/H| = a_{11}\cdots a_{nn}$. Let $A = (a_{ij})$. Let $\alpha$ be the column vector $(\alpha_1,\cdots,\alpha_n)^t$. Then $\alpha = A\omega$. Since $\{\alpha_1,\cdots,\alpha_n\}$ and $\{\beta_1,\cdots,\beta_n\}$ are free bases of $H$, there exists a matrx $P \in GL_n(\mathbb{Z})$ such that $\beta = P\alpha$. Then $P\alpha = B\omega$. Hence $PA\omega = B\omega$. Hence $PA = B$. Hence det $B =$ det $P$ det $A$. Since det $P = \pm 1$, |det $B| = |$det $A| = a_{11}\cdots a_{nn} = |G/H|$. QED