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Let $A,B$ be abelian groups such that $B\subseteq A$ and $A,B$ both are free of rank $n$. I want to show that $|A/B|$ is finite, or equivalently that $[A:B ]$ (the index of $B$ in $A$) is finite.

For example, if $A=\mathbb{Z}^n$ and $B=(2\mathbb{Z})^n$ we have that $[A:B]=2^n$ (correct me if I'm wrong).

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up vote 10 down vote accepted

There are several ways - for example induction on $n$ should work. We can also use the stacked basis theorem, if so desired.

But let's do the following. Select a basis ${\mathcal B}_A =\{a_1,a_2,\ldots,a_n\}$ of $A$. Let us similarly select a basis ${\mathcal B}_B=\{b_i,b_2,\ldots, b_n\}$ of $B$. As $B\subseteq A$ we get a matrix $M=(m_{ij})$, with integer entries, uniquely determined by $$ b_i=\sum_j m_{ij}a_i. $$ I want to show that the matrix $M$ is non-singular. Let $V$ be the vector space over $\mathbf{Q}$ with basis ${\mathcal B}_A$. Then ${\mathcal B}_A$ and ${\mathcal B}_B$ are both linearly independent over $\mathbf{Z}$ in $V$, because they were so in $A$. Thus they are both vector space bases of $V$, and $M$ is the change of bases matrix that is then non-singular.

The inverse matrix $M^{-1}$ then has the property that $(\det M)M^{-1}$ also has integer entries (this follows immediately from Cramer's rule). This means that (working inside $V$ again) the basis elements $a_i$ have the property that $(\det M)a_i$ is a linear combination of vectors $b_j$ with integer coefficients. Therefore $a_i\in(\det M)B$. As this holds for all $a_i$, we have shown that $$ (\det M)A\subseteq B. $$ The claim follows from this: $A/B$ is a finitely generated abelian group with exponent that is a factor of $|\det M|$, so it must be finite.

A more careful study (using stacked bases theorem) reveals that in fact we have $$ [A:B]=|\det M|. $$

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In the third paragraph, what is ${\mathcal B}$? – Carlos I. Jan 17 at 2:06
@CarlosI. A subscript was missing. Sorry about that. – Jyrki Lahtonen Jan 17 at 6:17

Let $A\cong B\cong \mathbb{Z}^n$, with $B\subseteq A$. Then we have the short exact sequence $$ 0\rightarrow B\rightarrow A\rightarrow A/B\rightarrow 0.$$

Tensoring with $\mathbb{Q}$ shows $A/B$ has rank $0$, so is torsion.

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I mention a slightly different approach to Jyrki's, though essentially equivalent, using the theory of finitely generated torsion free modules over a principal ideal domain (every finitely generated free Abelian group is such a module over the pid $\mathbb{Z}).$ If $A$ is a finitely generated module over a PID $R$, and $B$ is a submodule of $A,$ then there is a basis $\{a_{i}: 1 \leq i \leq m \}$ of $A$ and there are elements $d_{1},d_{2},\ldots d_{m}$ of $R$ such that $d_{i} | d_{i+1}$ in $R$ for each $i$ and $\{ d_{i}a_{i}: 1 \leq i \leq m \}$ is a generating set for $B$ (since we did not assume that $B$ had a basis of the same size as $A$, some of the $d_i$ may be $0,$ which is why I did not say it was a basis for $B).$ In the case $R = \mathbb{Z}$, this links with Jyrki's answer because the $d_{i}$'s are the invariant factors of the associated matrix, and $[A:B] = \prod_{i=1}^{m}d_{i}$ when $A$ and $B$ have the same rank. This theorem about $\mathbb{Z}$-modules leads to the Smith normal form for integral matrices.

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Perhaps the theorem I quote is what Jyrki is referring to as the "stacked basis theorem"-the name is unfamiliar to me. – Geoff Robinson Jun 10 '12 at 12:35
Geoff, that's what it was called by classmates and I in grad school. I think the same name was used in one of the standard textbooks used in U.S. grad schools. I didn't find it in Jacobson, but my search was not particularly thorough. – Jyrki Lahtonen Jun 12 '12 at 12:10
Yes, OK, I thought something like that might be the case. I know the theorem, just not that name for it. – Geoff Robinson Jun 12 '12 at 12:14

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