Finitely Generated Abelian Groups

Consider an abelian group \(A\) generated by \(m\) elements \[ A = \langle a_1,…​,a_m \rangle \] Then the free abelian group of rank \(m\) \[ F = \langle u_1,…​,u_m \rangle \] 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 \[ A = \mathbb{Z}^r \oplus \mathbb{Z}_{h_1} \oplus …​ \oplus \mathbb{Z}_{h_n} \] 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 \[ \array { A = \mathbb{Z}^r \oplus \mathbb{Z}_{e_1} \oplus …​ \oplus \mathbb{Z}_{e_n} \\ A = \mathbb{Z}^s \oplus \mathbb{Z}_{d_1} \oplus …​ \oplus \mathbb{Z}_{d_m} } \] 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 \[ \array { T = \mathbb{Z}_{e_1} \oplus …​ \oplus \mathbb{Z}_{e_n} \\ T = \mathbb{Z}_{d_1} \oplus …​ \oplus \mathbb{Z}_{d_m} } \] 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 \[ \langle x \rangle = \langle (n/p_1^{a_1}) x \rangle \oplus \langle (n/p_2^{a_2}) x \rangle \oplus …​ \]

Now let \[ e_1 = p_1^{a_1} p_2^{a_2} …​, e_2 = p_1^{b_1} p_2^{b_2} …​, …​ \] where \(a_i \le b_i \le …​\) for all \(i\) since \(e_i | e_i+1\). Then we have \[ T = P_1 \oplus P_2 \oplus …​ \oplus Q_1 \oplus Q_2 \oplus …​ \] 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 \[ \langle x,y \rangle = \langle x y \rangle \] 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 \[ T = P_1 \oplus P_2 \oplus …​ \oplus Q_1 \oplus Q_2 \oplus …​ \] we deduce that \[ T = \mathbb{Z}_{e_1} \oplus …​ \oplus \mathbb{Z}_{e_n} \] 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 \[ \array { A &=& \langle u_1 \rangle \oplus …​ \oplus \langle u_k \rangle \\ A &=& \langle v_1 \rangle \oplus …​ \oplus \langle v_l \rangle } \] where \(u_1,…​,u_k\) have orders \(p^{f_1} \ge …​ \ge p^{f_k} > 1\), and \(v_1,…​,v_v\) have orders \(p^{g_1} \ge …​ \ge p^{g_l} > 1\). 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 \[ \array { A_p &=& \langle p^{f_1-1} u_1 \rangle \oplus …​ \oplus \langle p^{f_k-1} u_k \rangle \\ A_p &=& \langle p^{g_1-1} v_1 \rangle \oplus …​ \oplus \langle p^{g_l-1} v_l \rangle } \] 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 \[ A^p = \langle p u_1 \rangle + …​ + \langle p u_\kappa \rangle = \langle p v_1 \rangle + …​ + \langle p v_\lambda \rangle \] 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 \[ A \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \cong (\mathbb{Z}_4 \oplus \mathbb{Z}_2) \oplus (\mathbb{Z}_3 \oplus \mathbb{Z}_3) \oplus \mathbb{Z}_5 \] 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 \[ A \cong F/R \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_2 \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_6 \]