Abelian Groups

Let $(G, +)$ denote an Abelian group with additive identity $0$ .

Definition 1. We say that $G$ is finitely generated if there exists a set $\{x_1,\dots, x_n\} \subseteq G$ , called a generating set that spans $G$ , such that

$G = \langle x_1 \rangle + \cdots + \langle x_n \rangle.$

For any $x \in G$ and $n \in \mathbb Z$ , define the map

$(n,x) = \begin{cases} 0, & n = 0, \\ \underbrace{x+\cdots+x}_n, & n > 0, \\ (-n, x), & n < 0. \end{cases}$

Denote $nx = (n,x)$ for brevity. It is not hard to check that $(-n)x = -(nx)$ for any integer $n$ . Furthermore,

$\langle x \rangle = \{nx : n \in \mathbb Z\} \equiv \mathbb Z\{x\}.$

More generally, the set $\{x_1,\dots, x_n\} \subseteq G$ is linearly independent if for integers $c_1,\dots, c_n$ ,

$\displaystyle \sum_{i=1}^n v_i x_i = 0 \quad \Rightarrow \quad v_i = 0.$

A generating set for $G$ that is linearly independent is called a basis for $G$ .

Definition 2. We say that the Abelian group $G$ is free if there exists a positive integer $n$ such that $G \cong \mathbb Z^n$ . Here, the right-hand side is a direct product in the following sense: letting $\odot_i$ denote the binary operation in the set $K_i$ , define $\odot$ on $K_1\times \cdots \times K_n$ by

$\begin{bmatrix} u_1 \\ \vdots \\ u_n \end{bmatrix} \odot \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix} := \begin{bmatrix} u_1 \odot_1 v_1 \\ \vdots \\ u_n \odot_n v_n \end{bmatrix}.$

More precisely then, the isomorphism means $(G, + ) \cong (\mathbb Z^n, +)$ . In this case, we call $n$ the rank of $G$ .

Problem 1. Suppose $G$ is free. Show that $G$ is finitely generated by a basis $\{x_1,\dots, x_n\}$ , and

$\displaystyle G \cong \langle x_1 \rangle \oplus \cdots \oplus \langle x_n \rangle.$

(Click for Solution)

Solution. Since $G$ is free, let $\phi : \mathbb Z^n \to G$ denote the group isomorphism. Define $x_i := \phi(\mathbf e_i)$ . We claim that $\{x_1,\dots, x_n\}$ forms the desired basis.

For any $x \in G$ , there exists $\mathbf v \in \mathbb Z^n$ such that $x = \phi(\mathbf v)$ . By definition, there exists integers $v_i$ such that $\mathbf v = \sum_{i=1}^n v_i \mathbf e_i$ . Since $\phi$ is a homomorphism,

$\displaystyle x = \phi(\mathbf v) = \sum_{i=1}^n v_i \phi(\mathbf e_i) = \sum_{i=1}^n v_i x_i.$

Linear independence follows suit, since the isomorphism implies that

$\displaystyle \sum_{i=1}^n v_i x_i = 0 \quad \Rightarrow \quad \sum_{i=1}^n v_i \mathbf e_i = \mathbf 0,$

so that $v_i = 0$ for each $i$ . For the direct product claim, we check that the map

$\displaystyle (v_1 x_1, \dots, v_n x_n) \mapsto \sum_{i=1}^n v_i x_i$

is an isomorphism similar as before.

Remark 1. The spanning property holds as long as $\phi$ is surjective, while the linear independence property holds as long as $\phi$ is injective.

Problem 2. Suppose $G$ is free and $H \leq G$ . Show the following:

$H$ is free of rank $m \leq n$ .
There exists a basis $\{u_1,\dots, u_n\}$ for $G$ and positive integers $c_1,\dots ,c_m$ such that $\{c_1u_1,\dots, c_mu_m\}$ forms a basis for $H$ .
$[G : H]$ is finite if and only if $m = n$ .

(Click for Solution)

Solution. Let $G$ be free with rank $n$ . Let $\phi : \mathbb Z^n \to G$ denote the given isomorphism. Since $H \leq G$ , $\phi^{-1}(H) \leq \mathbb Z^n$ .

It suffices to assume $G = \mathbb Z^n$ . We prove by induction on $n$ . For $n = 1$ , $H \leq \mathbb Z$ implies that

$H = \alpha \mathbb Z = \langle \alpha \rangle \cong \mathbb Z$

for some $\alpha \in \mathbb Z^+$ . Then $\{\alpha\}$ serves as the desired basis for $H$ .

Suppose the result holds for free subgroups with rank $< n$ . Let $\pi : \mathbb Z^n \to \mathbb Z$ denote projection onto the last coordinate, i.e. $(v_1,\dots,v_n)^{\mathrm T} \mapsto v_n$ . It is clear that $\pi$ is a surjective linear transformation, even when restricted to $\mathbb Z$ as the underlying set of scalars. Then $\ker(\pi) \cong \mathbb Z^{n-1}$ via the identification

$(v_1,\dots, v_{n-1},0) \cong (v_1,\dots, v_{n-1}).$

Since $\ker(\pi) \leq G$ and $H \leq G$ ,

$H \cap \ker(\pi) \leq \ker(\pi) \cong \mathbb Z^{n-1}.$

By induction, $H \cap \ker(\phi) \cong \mathbb Z^k$ with $k \leq n-1$ . Similarly, $\pi(H) \cong \mathbb Z^\ell$ is free with $\ell \leq 1$ . Check that

$H \cong (H \cap \ker(\pi)) \times \pi(H),$

so that

$\begin{aligned} H \cong (H \cap \ker(\pi)) \times \pi(H) &\cong \mathbb Z^k \times \mathbb Z^\ell \\ &\cong \mathbb Z^{k+\ell} \\ &\leq \mathbb Z^n, \end{aligned}$

as required.

Now we prove the second two propositions. Let $\mathbf x := (x_1,\dots, x_n)^{\mathrm T}$ be an (ordered) basis for $G$ and $\mathbf y := (y_1, \dots, y_m)^{\mathrm T}$ be an (ordered) basis for $H$ . By writing each $y_i$ in terms of the $x_j$ , we can obtain a matrix $\mathbf A \in \mathcal M_{m \times n}(\mathbb Z)$ such that

$\mathbf y = \mathbf A \mathbf x.$

Let $\mathbf M$ denote the invertible matrix capturing the process of reducing an augmented matrix into reduced row-echelon form (RREF), so that $\mathbf R = \mathbf M \mathbf A$ denotes the RREF of $\mathbf A$ . Then

$\mathbf M \mathbf y = \mathbf R \mathbf x.$

Then the elements $\{y_1,\dots, y_m\}$ that correspond to the $m$ pivot-columns in $\mathbb R$ yields the desired basis for $H$ . If $m < n$ , then $\mathbb Z \leq G/H$ , and $[G : H] = \infty$ .

Problem 3. Now suppose instead that $G$ is finitely generated. Show that there exists an integer $r \geq 0$ , primes $p_s$ , and integers $\alpha_s > 0$ such that

$\displaystyle G \cong \mathbb Z^r \oplus \bigoplus_{s=1}^k \mathbb Z/p_s^{\alpha_s} \mathbb Z,$

where $\mathbb Z/n\mathbb Z = \mathbb Z_n$ by definition. This result is one version of the fundamental theorem of finitely generated Abelian groups.

(Click for Solution)

Solution. Let $\{x_1,\dots, x_n\}$ denote a basis for $G$ and $\mathcal I$ denote the collection of indices such that $|\langle x_i \rangle| = \infty$ . For $i \in \mathcal I$ , $\langle x_i \rangle \cong \mathbb Z$ . For $j \in \{1,\dots, n\} \backslash \mathcal I =: \mathcal J$ , there exists $k_j \in \mathbb Z^+$ such that $\langle x_j \rangle \cong \mathbb Z/k_j \mathbb Z$ . Since

$\displaystyle G \cong \bigoplus_{i \in \mathcal I} \langle x_i \rangle \oplus \bigoplus_{j \in \mathcal J} \langle x_j \rangle,$

intuitively, we would like the first piece to be isomorphic to $\mathbb Z^{|\mathcal I|}$ , and the second to be isomorphic to a product of $\mathbb Z/k_j\mathbb Z$ . If $\mathcal J = \emptyset$ , then $G \cong \mathbb Z^n$ trivially. Otherwise, define

$\displaystyle H := \bigoplus_{j \in \mathcal J} \langle x_j \rangle \leq G.$

By Problem 2, $H$ is free of rank $m := |\mathcal J| \leq n$ , and finite. Hence, there exists a basis $\{u_1,\dots, u_m\}$ for $G$ and positive integers $c_1,\dots, c_m$ such that $\{c_1u_1,\dots, c_mu_m\}$ forms a basis for $H$ . In particular,

$\begin{aligned} G \cong \bigoplus_{r=1}^m \langle u_r \rangle \oplus \bigoplus_{r=m+1}^n \langle u_r \rangle,\quad H \cong \bigoplus_{r=1}^m c_r \langle u_r \rangle. \end{aligned}$

Then, the required isomorphism

$\displaystyle \phi : G \to \bigoplus_{r=1}^m \mathbb Z/c_r \mathbb Z \times \mathbb Z^{n-m}.$

is defined by

$\displaystyle \sum_{r=1}^n a_r u_r \mapsto ([a_1]_{c_1},\dots, [a_m]_{c_m}, a_{m+1} , \dots, a_n).$

where $[a]_c := a\ \text{mod}\ c$ . Since $\{c_1u_1,\dots, c_mu_m\}$ forms a basis, $\gcd(c_i,c_j) = 1$ for $i \neq j$ . By the fundamental theorem of arithmetic, there exists unique primes $p_i$ and integers $\alpha_i > 0$ such that