# Finitely Generated Abelian Groups

Consider an abelian group \(A\) generated by \(m\) elements

Then the free abelian group of rank \(m\)

maps homomorphically onto \(A\) via the map that sends \(u_i\) to \(a_i\). By the first isomorphism theorem we have \(A \cong F / R\) for some subgroup \(R\) of \(F\). Pick a basis \(v_1,...,v_m\) of \(F\) such that \(R = \langle h_1 v_1 ,..., h_q v_q \rangle\) where \(h_i | h_{i+1}\), \(h_i \ge 1\), \(q \le m\).

Consider the case where \(m=1\). There are three possibilities. (1) \(R = \langle v \rangle\), so \(F / R\) is the trivial group, (2) \(R = \langle h v \rangle\), in which case \(F/R = \mathbb{Z}_h\), and (3) \(R = \{ 0\}\) and we have \(F/R = F\).

More generally, we have:

**Theorem:** Every finitely generated abelian group can be expressed as
the direct sum of cyclic groups

where \(h_i | h_{i+1}\).

**Corollary:** A finitely generated abelian group is free if and only if
it is **torsion-free**, that is, it contains no element of finite order
other than the identity.

The number \(r\) is called the **rank** of \(A\). The orders of the cyclic
groups \(h_1,...,h_n\) are called the **invariants** of \(A\). Note \(A\) is
finite if and only if its rank is zero.

**Theorem:** Suppose \(A\) is a finitely generated abelian group with
decompositions

satifying \(e_i | e_{i+1}, d_i | d_{i+1}\). Then \(r=s, n=m, e_i=d_i\).

**Proof:** Let \(T\) be the set of elements of \(A\) of finite order.
Clearly if \(g,h\) have finite order then \(ord(g) ord(h) (h-k) = 0\) hence
\(h-k\) also has finite order hence \(T\) is a subgroup of \(A\). It is called
the **torsion group** of \(A\).

A little thought shows that we must have

Consider the map that projects \(A\) onto \(\mathbb{Z}^r\). By the first isomorphism theorem we have that \(A/T \cong \mathbb{Z}^r\). Similarly we have \(A/T \cong \mathbb{Z}^s\) hence \(r=s\).

Now conisder \(T\). Let \(p\) be a prime, and let \(P\) be the set of elements whose order is a power of \(P\). Then \(P\) is a group. We first need the following:

**Theorem:** Let \(G\) be a finite abelian group of order
\(p_1^{a_1} p_2^{a_2} ...\) where the \(p_i\)'s are distinct primes. Then
\(G = P_1 \oplus P_2 \oplus ...\) where \(P_i\) is the subgroup of elements whose
orders are powers of \(p_i\).

**Proof:**
Let \(x \in G\) be an element of order \(p_1^{\alpha} f_1\) where \(f_1, p_1\)
are coprime. Then we may write \(x = a_1 + x_1\)
where \(a_1\) has order \(p_1^{\alpha}\)
and \(x_1\) has order \(f_1\). (Simply take \(a_1 = u f_1 x, x_1 = v p_1^{\alpha} x\)
where \(u f_1 + v p_1^{\alpha} = 1\).)

Iterating this procedure gives a decomposition \(x = a_1 + a_2 + ...\) with \(a_i \in P_i\). We claim this decomposition is unique. Suppose \(0 = b_1 + b_2 + ...\) where \(b_i \in P_i\). Then for all \(i\), subtracting \(b_i\) from both sides shows that the order of \(b_i\) is coprime to \(p_i\). But it must also be a power of \(p_i\) which is only possible if \(b_i = 0\).

It is clear that the groups \(P_i\) are uniquely determined. In fact, they are the Sylow groups since \(G\) is abelian.∎

In particular, if \(x\) is an element of order \(n = p_1^{a_1} p_2^{a_2} ...\) then we have

Now let

where \(a_i \le b_i \le ...\) for all \(i\) since \(e_i | e_i+1\). Then we have

where \(P_1,P_2,...,Q_1,Q_2,...\) are cyclic groups of order \(p_1^{a_1}, p_1^{b_1}, ..., p_2^{a_2}, p_2^{b_2},...\). We see that the Sylow groups of \(T\) are \(P = P_1 \oplus P_2 \oplus ..., Q = Q_1 \oplus Q_2 \oplus ...\). Now we need the following:

**Lemma:** Let \(G\) be any group. Suppose \(x, y \in G\) commute and have
relatively prime orders \(m, n\). Then

is cyclic of order \(m n\).

**Proof:** We know the order is at most \(m n\) since each element must
be of the form \(x^a y^b\) for \(a=0,...,m-1, b = 0,...,n-1\). Now
suppose \((x y)^t = 1\). Then \(1 = (x y)^{t m} = y^{t m}\) implying
that \(n | t m\). Since \(m, n\) is coprime we have \(n | t\). Similarly
\(m | t\), thus the group order must be exactly \(m n\).∎

Thus given

we deduce that

so that one decomposition implies the other. We are done as soon as we show that the Sylow groups have a unique decomposition:

**Theorem:**
Let \(A\) be an abelian group of order \(p^a\) where \(p\) is prime.
Suppose

where \(u_1,...,u_k\) have orders \(p^{f_1} \ge ... \ge p^{f_k} \gt 1\), and \(v_1,...,v_v\) have orders \(p^{g_1} \ge ... \ge p^{g_l} \gt 1\). Then \(k =l\) and \(f_i = g_i\) for \(i = 1,...,k\).

**Proof:**
Note we must have \(a = f_1 + ... + f_k = g_1 + ... + g_l\). The theorem is
trivial when \(a = 1\), which we use to start an induction.

Let \(A_p\) be the set of elements \(x \in A\) with \(p x = 0\). Then \(A_p\) is a subgroup. We have

Hence \(A_p = \mathbb{Z}_p^k = \mathbb{Z}_p^l\) implying that \(k=l\).

Now consider the set \(A^p\) of elements \(p x\) for all \(x \in A\) (the multiples of \(p\)). Then \(A^p\) is a subgroup, and is generated by \(p u_1 , ...,p u_k\) and also by \(p v_1 ,..., p v_k\). But in general these are not bases for \(A^p\) since we might have \(p u_i = 0\) for example. So find \(\kappa\) such that \(f_1,...,f_\kappa \ge 2\) and \(f_{\kappa+1} = ... = f_k = 1\), and similarly find \(\lambda\) with \(g_1,...,g_\lambda \ge 2\) and \(g_{\lambda+1} = ... = g_k = 1\).

This yields the decompositions

By inductive hypothesis we have \(\kappa = \lambda\) and \(f_i - 1 = g_i - 1\) for all \(i = 1,...,\kappa\).∎

We have now proved the main theorem.∎

In the last proof, the numbers \(p^{f_1},...,p^{f_k}\) are called the
*elementary divisors* of \(A\) corresponding to \(p\).
\(A\) is said to be of *type* \((f_1,...,f_k)\).

**Example:** Suppose an abelian group \(A\) is generated by \(a,b\) subject
to the relations \(30 a = 12 b = 0\). Then define the free abelian groups
\(F = \langle x,y \rangle\) and \(R = \langle 30x, 12 y \rangle\). Note
we have \(A \cong F / R = \mathbb{Z}_{30} \oplus \mathbb{Z}_{12}\).
Then we have

Thus the elementary divisors for 2,3,5 are \((4,2), (3,3), 5\). Rearranging gives \(A \cong \mathbb{Z}_{60} \oplus \mathbb{Z}_6\), so the invariants are \(60, 6\).

**Example:** Suppose an abelian group \(A\) is generated by
\(a,b,c,d\) and the relations \(3a + 9b -3c =0, 4a+2b-2d=0\). Then
define the free abelian groups \(F=\langle x,y,z,t\rangle\) and
\(R=\langle 3u,2v\rangle\) where \(u=x+3y-z, v=2x +y -t\). Note \(x,y,u,v\) is
also a basis of \(F\). Thus

*blynn@cs.stanford.edu*💡