Finitely generated abelian groups

We will now prove the structure theorem for finitely generated abelian groups, since it will be crucial for much of what we will do later.

Let $ \mathbf{Z}=\{0,\pm 1, \pm 2, \ldots\}$ denote the ring of integers, and for each positive integer $ n$ let $ \mathbf{Z}/n\mathbf{Z}$ denote the ring of integers modulo $ n$, which is a cyclic abelian group of order $ n$ under addition.

Definition 3.0.1 (Finitely Generated)   A group $ G$ is if there exists $ g_1,\ldots,
g_n \in G$ such that every element of $ G$ can be obtained from the $ g_i$.

Theorem 3.0.2 (Structure Theorem for Abelian Groups)   Let $ G$ be a finitely generated abelian group. Then there is an isomorphism

$\displaystyle G \cong (\mathbf{Z}/n_1\mathbf{Z}) \oplus (\mathbf{Z}/n_2\mathbf{Z}) \oplus \cdots \oplus
(\mathbf{Z}/n_s\mathbf{Z}) \oplus \mathbf{Z}^{r},

where $ n_1>1$ and $ n_1\mid{}n_2\mid{}\cdots \mid{}n_s$. Furthermore, the $ n_i$ and $ r$ are uniquely determined by $ G$.

We will prove the theorem as follows. We first remark that any subgroup of a finitely generated free abelian group is finitely generated. Then we see that finitely generated abelian groups can be presented as quotients of finite rank free abelian groups, and such a presentation can be reinterpreted in terms of matrices over the integers. Next we describe how to use row and column operations over the integers to show that every matrix over the integers is equivalent to one in a canonical diagonal form, called the Smith normal form. We obtain a proof of the theorem by reinterpreting in terms of groups.

Proposition 3.0.3   Suppose $ G$ is a free abelian group of finite rank $ n$, and $ H$ is a subgroup of $ G$. Then $ H$ is a free abelian group generated by at most $ n$ elements.

The key reason that this is true is that $ G$ is a finitely generated module over the principal ideal domain $ \mathbf{Z}$. We will give a complete proof of a beautiful generalization of this result in the context of Noetherian rings next time, but will not prove this proposition here.

Corollary 3.0.4   Suppose $ G$ is a finitely generated abelian group. Then there are finitely generated free abelian groups $ F_1$ and $ F_2$ such that $ G \cong F_1/F_2$.

Proof. Let $ x_1,\ldots, x_m$ be generators for $ G$. Let $ F_1=\mathbf{Z}^m$ and let $ \varphi :F_1\to G$ be the map that sends the $ i$th generator $ (0,0,\ldots,1,\ldots,0)$ of $ \mathbf{Z}^m$ to $ x_i$. Then $ \varphi $ is a surjective homomorphism, and by Proposition 3.0.4 the kernel $ F_2$ of $ \varphi $ is a finitely generated free abelian group. This proves the corollary. $ \qedsymbol$

Suppose $ G$ is a nonzero finitely generated abelian group. By the corollary, there are free abelian groups $ F_1$ and $ F_2$ such that $ G \cong F_1/F_2$. Choosing a basis for $ F_1$, we obtain an isomorphism $ F_1\cong \mathbf{Z}^n$, for some positive integer $ n$. By Proposition 3.0.4, $ F_2\cong \mathbf{Z}^m$, for some integer $ m$ with $ 0\leq m\leq n$, and the inclusion map $ F_2\hookrightarrow F_1$ induces a map $ \mathbf{Z}^m\to \mathbf{Z}^n$. This homomorphism is left multiplication by the $ n\times m$ matrix $ A$ whose columns are the images of the generators of $ F_2$ in $ \mathbf{Z}^n$. The of this homomorphism is the quotient of $ \mathbf{Z}^n$ by the image of $ A$, and the cokernel is isomorphic to $ G$. By augmenting $ A$ with zero columns on the right we obtain a square $ n\times n$ matrix $ A$ with the same cokernel. The following proposition implies that we may choose bases such that the matrix $ A$ is diagonal, and then the structure of the cokernel of $ A$ will be easy to understand.

Proposition 3.0.5 (Smith normal form)   Suppose $ A$ is an $ n\times n$ integer matrix. Then there exist invertible integer matrices $ P$ and $ Q$ such that $ A'=PAQ$ is a diagonal matrix with entries $ n_1, n_2,\ldots, n_s,0,\ldots,0$, where $ n_1>1$ and $ n_1\mid n_2 \mid{} \ldots \mid{} n_s$. This is called the Smith normal form of $ A$.

We will see in the proof of Theorem 3.0.3 that $ A'$ is uniquely determined by $ A$.

Proof. The matrix $ P$ will be a product of matrices that define elementary row operations and $ Q$ will be a product corresponding to elementary column operations. The elementary operations are:
  1. Add an integer multiple of one row to another (or a multiple of one column to another).
  2. Interchange two rows or two columns.
  3. Multiply a row by $ -1$.
Each of these operations is given by left or right multiplying by an invertible matrix $ E$ with integer entries, where $ E$ is the result of applying the given operation to the identity matrix, and $ E$ is invertible because each operation can be reversed using another row or column operation over the integers.

To see that the proposition must be true, assume $ A\neq 0$ and perform the following steps (compare [Art91, pg. 459]):

  1. By permuting rows and columns, move a nonzero entry of $ A$ with smallest absolute value to the upper left corner of $ A$. Now attempt to make all other entries in the first row and column 0 by adding multiples of row or column 1 to other rows (see step 2 below). If an operation produces a nonzero entry in the matrix with absolute value smaller than $ \vert a_{11}\vert$, start the process over by permuting rows and columns to move that entry to the upper left corner of $ A$. Since the integers $ \vert a_{11}\vert$ are a decreasing sequence of positive integers, we will not have to move an entry to the upper left corner infinitely often.

  2. Suppose $ a_{i1}$ is a nonzero entry in the first column, with $ i>1$. Using the division algorithm, write $ a_{i1} = a_{11}q + r$, with $ 0\leq r < a_{11}$. Now add $ -q$ times the first row to the $ i$th row. If $ r>0$, then go to step 1 (so that an entry with absolute value at most $ r$ is the upper left corner). Since we will only perform step 1 finitely many times, we may assume $ r=0$. Repeating this procedure we set all entries in the first column (except $ a_{11}$) to 0. A similar process using column operations sets each entry in the first row (except $ a_{11}$) to 0.

  3. We may now assume that $ a_{11}$ is the only nonzero entry in the first row and column. If some entry $ a_{ij}$ of $ A$ is not divisible by $ a_{11}$, add the column of $ A$ containing $ a_{ij}$ to the first column, thus producing an entry in the first column that is nonzero. When we perform step 2, the remainder $ r$ will be greater than 0. Permuting rows and columns results in a smaller $ \vert a_{11}\vert$. Since $ \vert a_{11}\vert$ can only shrink finitely many times, eventually we will get to a point where every $ a_{ij}$ is divisible by $ a_{11}$. If $ a_{11}$ is negative, multiple the first row by $ -1$.
After performing the above operations, the first row and column of $ A$ are zero except for $ a_{11}$ which is positive and divides all other entries of $ A$. We repeat the above steps for the matrix $ B$ obtained from $ A$ by deleting the first row and column. The upper left entry of the resulting matrix will be divisible by $ a_{11}$, since every entry of $ B$ is. Repeating the argument inductively proves the proposition. $ \qedsymbol$

Example 3.0.6   The matrix $ \left(
\begin{matrix}1&2\ 3&4
\end{matrix}\right)$ is equivalent to $ \left(
\begin{matrix}1&0\ 0&2
\end{matrix}\right)$ and the matrix $ \left(
\begin{matrix}1&2&3\ 4&5&6\ 7&8&9
\end{matrix}\right)$ is equivalent to $ \left(
\begin{matrix}1&0&0\ 0&3&0\ 0&0&0
\end{matrix}\right).$ Note that the determinants match, up to sign.

Proof. [Theorem 3.0.3] Suppose $ G$ is a finitely generated abelian group, which we may assume is nonzero. As in the paragraph before Proposition 3.0.6, we use Corollary 3.0.5 to write $ G$ as a the cokernel of an $ n\times n$ integer matrix $ A$. By Proposition 3.0.6 there are isomorphisms $ Q:\mathbf{Z}^n\to \mathbf{Z}^n$ and $ P:\mathbf{Z}^n\to \mathbf{Z}^n$ such that $ A'=PAQ$ is a diagonal matrix with entries $ n_1, n_2,\ldots, n_s,0,\ldots,0$, where $ n_1>1$ and $ n_1\mid n_2 \mid{} \ldots \mid{} n_s$. Then $ G$ is isomorphic to the cokernel of the diagonal matrix $ A'$, so

$\displaystyle G \cong (\mathbf{Z}/n_1\mathbf{Z}) \oplus (\mathbf{Z}/n_2\mathbf{Z}) \oplus \cdots \oplus
(\mathbf{Z}/n_s\mathbf{Z}) \oplus \mathbf{Z}^{r},
$ (3.1)

as claimed. The $ n_i$ are determined by $ G$, because $ n_i$ is the smallest positive integer $ n$ such that $ nG$ requires at most $ s+r-i$ generators (we see from the representation (3.0.1) of $ G$ as a product that $ n_i$ has this property and that no smaller positive integer does).

$ \qedsymbol$

William Stein 2004-05-06