## 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