# 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

$A = \mathbb{Z}^n \oplus \mathbb{Z}_{h_1} \oplus ... \oplus \mathbb{Z}_{h_n}$

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 $$p^{f_1} \ge ... \ge p^{f_k} \gt 1$$, 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

$\array { A_p &=& \langle p^{f_1-1} u_1 \rangle \oplus ... \oplus \langle p^{f_k-1} u_k \rangle \\ A_p &=& \langle p^{g_1-1} v_1 \rangle \oplus ... \oplus \langle p^{g_l-1} v_l \rangle }$

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

$A \cong F/R \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_2 \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_6$

Ben Lynn blynn@cs.stanford.edu 💡