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In Alperin's book, Local Representation Theory (p. 169) there is a claim I am finding difficult to verify.

The setup is the following. Given a free abelian group $V$ spanned by basis elements $\{v_1, \ldots, v_n\}$ and $n$ other elements $\{w_1, \ldots, w_n\}$ spanning subgroup $W$, we want to calculate the size (in number of elements) of $V/W$.

The claim is that this is finite iff the matrix $C$ expressing the $w_i$ in terms of the $v_i$ has nonzero determinant. Moreover if this is so, its exact size is the modulus of the determinant of $C$.

This is explained only as "by the theory of elementary divisors" and I was wondering if anybody had either a proof of this, or directions to some material towards a proof of this.

Edit: specified "...exact size is the modulus of the determinant..."

rspencer
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  • The "theory of elementary divisors" is purely linear algebra : see e.g., this. I have no idea about its application to group theory. – Jean Marie Mar 25 '18 at 21:20
  • If the determinant is zero, then the span of $W$ over the rationals has dimension less than $n$ as a vector space over the rationals. If $v$ is an element of the group not in the span of $W$, then the element (coset) $v + W$ has infinite order in the quotient group. Conversely, if the determinant is nonzero, then a similar argument shows each coset $v_i +W$ has finite order (since $v_i$ together with the collection of $w_j$ is linearly dependent and you can clear denominators) in the quotient. I don't see how to prove that the order is correct, but geometrically this seems right as a volume – Robert Bell Mar 25 '18 at 23:25
  • See also https://math.stackexchange.com/questions/245733/index-of-a-sublattice-in-a-lattice-and-a-homomorphism-between-them – lhf Mar 27 '18 at 13:28

2 Answers2

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Invariant factors refers to Smith normal form (link: wikipedia).

Write $w_i = \sum_{j = 1}^n A_{ji} v_j$, where $A_{ji} \in \mathbb{Z}$.

The idea is that by changing the generators of $W$ and $V$ suitably, we can find new generating sets $\{ w_1', \ldots, w_n' \}$ of $W$ and $\{v_1', \ldots, v_n'\}$ of $V$ such that for all $i$, $$w_i' = k_i v_i'$$ for some non-negative integers $k_i$ such that $k_i \mid k_{i+1}$. This follows from the Smith normal form applied to the integer matrix $A$. Here we have $|\det A| = k_1 \cdots k_n$.

It is obvious from $w_i' = k_iv_i'$ that $$V/W \cong \mathbb{Z}/k_1 \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/k_n\mathbb{Z}.$$

Hence $V/W$ is finite if and only if $k_i > 0$ for all $i$; so $V/W$ is finite if and only if $\det A \neq 0$. Finally, when $V/W$ is finite, it has order $k_1 \cdots k_n = |\det A|$.

spin
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    The Smith Normal form does not preserve determinants; they may be multiplied by units of the ring in question. The original question ought to have said "modulus of the determinant". – ancient mathematician Mar 26 '18 at 11:18
  • Good spot. Corrected. – rspencer Mar 26 '18 at 13:18
  • @ancientmathematician: Thanks, I edited my answer. Here in the case of $\mathbb{Z}$ the units are $\pm 1$, so up to a sign the order is given by the determinant. – spin Mar 27 '18 at 10:16
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$\mathcal{W} = C \mathcal{V}$ implies $\operatorname{adj}(C)\mathcal{W} = \det(C) \mathcal{V}$. Therefore, $\det(C)v \in W$ for every $v \in V$ and so $\det(C)$ kills every element of $V/W$. Thus, if $\det(C)\ne0$, then $V/W$ is a finitely generated torsion abelian group, and so must be finite.

lhf
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    I'd like to see a simple proof that the size of $V/W$ is exactly $\det(C)$ that does not use the Smith normal form. – lhf Mar 26 '18 at 10:40
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    @Ihf there are a couple geometric proofs at the end of page 10 and beginning of page 11 here: http://www.pdmi.ras.ru/~lowdimma/sandpile/sandpilelectures.pdf – markasoftware Jun 01 '21 at 05:37