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Let $G,H$ be free abelian groups such that $H\leq G$ and $rank(G)=rank(H)=n$.

Let $\alpha,\beta$ be bases for $G,H$ respectively and $M\in Mat_{n\times n}(\mathbb{Z})$ be the base changing matrix from $\beta$ to $\alpha$. (i.e. $M=[id]_\beta^{\alpha}$)

How do I prove that $|G:H|$ is finite and $|G:H|=|det(M)|$?

Rubertos
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1 Answers1

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One of the standard ways is to use Smith Normal Form. There are matrices $U$, $V$ over $\Bbb Z$ with determinant $\pm1$ such that $UMV$ is diagonal, with diagonal entries $d_1,\ldots,d_n$ say. Then $G/H \cong \prod_{i=1}^n(\Bbb Z/d_i\Bbb Z)$. As $G/H$ is finite, then all $d_i$ are finite, and $|G/H|=\prod_i |d_i|=|\det M|$.

Angina Seng
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