## Abelian Groups

We no longer assume that the groups we study are finite.

With abelian groups, additive notation is often used instead of multiplicative notation. In other words the identity is represented by $0$, and $a + b$ represents the element obtained from applying the group operation to $a$ and $b$.

A group $G$ is the direct sum of two subgroups $U, V$ if every element $x \in G$ can be written in the form $x = u + v$ where $u \in U, v \in V$, and $u + v = 0$ implies $u = v = 0$. We write $G = U \oplus V$.

Note that $U,V$ cannot have a nonzero element $w$ in common, otherwise $w + ({-w}) = 0$ is a nontrivial decomposition of zero. Also $u,v$ are uniquely determined by $x$ for if $u_1 + v_1 = u_2 + v_2$ implies $u_1 - u_2 = v_2 - v_1 \in U \cap V$.

More generally we have $G = U_1 \oplus ... \oplus U_r$, if every $x\in G$ can be written in the form $x = u_1 + ... + u_r$ and also if $0 = u_1 + ... + u_r$ implies $0 = u_1 = ... = u_r$. Clearly if $G$ is finite we have $|G| = |U_1| ... |U_r|$.

An abelian group $A$ is a free abelian group of rank $r$ if there exist $u_1,...,u_r \in A$ such that $A = \langle u_1,...,.u_r \rangle$ and $a_1 u_1 + ... + a_r u_r$ implies $a_1 = ... = a_r = 0$. Alternatively we may require every $x\in A$ can be uniquely written in the form $x =a_1 u_1 + ... + a_r u_r$. The set $\{u_1,...,u_r\}$ is a set of free generators of $A$. The trivial group is viewed as a free abelian group of rank zero, and viewed as been generated by the empty set.

Generators need not be unique. However it is easy to see that two sets of free generators are related by a unimodular (determinant of absolute value one) matrix transformation.

Theorem: [Dedekind] Let $F$ be a free abelian group of rank $r$ and let $G$ be a nonzero subgroup of $F$. Then $G$ is a free abelian group of rank $s$ with $s \le r$. Furthermore, $F$ has a set of free generators $\{u_1 ,..., u_r\}$ such that $G$ is generated by

$\array { v_1 &=& a_{1 1} u_1 &+& a_{1 2} u_2 & + &... &+& a_{1 r} u_r \\ v_2 &=& & & a_{2 2} u_2 & + &... &+ &a_{2 r} u_r \\ &\vdots& \\ v_s &=& & & & & a_{s s}u_s + ... &+& a_{s r} u_r }$

for some $a_{i j}$ with $a_{1 1}, a_{2 2},...,a_{s s}$ positive.

Proof: Let $\{u_1,...,u_r\}$ be free generators for $F$. Then take any nonzero element $b = b_1 u_1 + ... + b_r u_r$ of $G$. After permuting the $u_i$'s if necessary, assume $b_1 \ne 0$. Then since $G$ is closed under inverses, we may take $b_1 \gt 0$.

Enumerate all elements $x_1 u_1 + ... + x_r u_r$ of $G$ and consider the set of possible positive integer values for $x_1$. We know this set is nonempty since $b_1$ is a possible value. Then call the smallest integer in this set $a_{1 1}$ and take any element $v_1 = a_{1 1} u_1 + ... + a_{1 r} u_r \in G$ for which this minimum is attained.

Then every element $x_1 u_1 +...+ x_r u_r \in G$ must satisfy $a_{1 1} | x_1$, since we have $x_1 = a_{1 1} q + b$ for integers $q,b$ with $0 \le b \lt a_{1 1}$ (which implies $x_1 = b$ for some element of $G$), and we have chosen $a_{1 1}$ to be minimal.

Thus for all $x\in G$, for some integer $q$ we have $x - q v_1 = b_2 u_2 + ... + b_r u_r$ for some $b_2,...,b_r$. If $r = 1$ then we are done since we have $F = \langle u_1 \rangle$, $G = \langle a_{1 1} u_1\rangle$.

We use induction. Suppose $r \gt 1$. Let $F_1=\langle u_2,...,u_r\rangle, G_1 = G\cap G_1$. Then $G_1$ is a subgroup of $F_1$ and by inductive hypothesis $G_1 = \langle v_2,..,v_s\rangle$ where $s\le r$ and

$\array { v_2 &=& a_{2 2} u_2 &+& a_{2 3} u_2 & + &... &+& a_{2 r} u_r \\ v_3 &=& & & a_{3 3} u_3 & + &... &+ &a_{3 r} u_r \\ &\vdots& \\ v_s &=& & & & & a_{s s}u_s + ... &+& a_{s r} u_r }$

with $a_{2 2}, ..., a_{s s}$ positive. We claim $v_1, ...,v_s$ generate $G$. We have already seen that for any $x\in G$, there exists some integer $q$ such that $x-q v_1 \in F_1$. Then $x-q v_1 \in G_1$, hence $G = \langle v_1,...,v_s\rangle$.

It remains to show that $v_1,...,v_s$ are independent. Suppose not, that is, there exists a nontrivial relation $c_1 v_1 + ... + c_s v_s =0$. We must have $c_1 \ne 0$ because by induction we cannot have a nontrivial relation between $v_2 ,..., v_s$. Expressing the $v_i$'s in terms of the $u_i$'s, we arrive at a nontrivial relation between the $u_i$'s since the coeffecient of $u_1$ is $c_1 a_{1 1} \ne 0$, a contradiction since the $u_i$'s are independent.∎

Now let $F = \langle u_1,...,u_r\rangle$ be an abelian free group of rank $r$. Recall any set of generators of $F$ is related to the $u_i$'s via a unimodular matrix transformation, hence such a generator $b_1 u_1 + ... + b_r u_r$ must have $gcd(b_1,...,b_r) = 1$. The converse is also true:

Lemma: Let $F = \langle u_1 ,..., u_r \rangle$. Let $v = b_1 u_1 + ... + b_r u_r$ with $gcd(b_1,...,b_r) = 1$. Then there exist $v_2,...,v_r \in F$ with $F = \langle v, v_2,...,v_r\rangle$.

Proof: Set $s = |b_1| + ... + |b_r|$. If $s=1$ then the result is trivial, since we have $v = \pm u_i$ for some $i$. We shall induct on $s$.

If $s\gt 1$ then at least two of the $b_i$'s are nonzero, and without loss of generality assume $b_1 \ge \b_2 \gt 0$. Then set $u'_1 = u_1, u'_2 = u_1 + u_2, u'_j = u_j$ for $j\ge 3$. Clearly $F = \langle u'_1,...,u'_r \rangle$, and we have

$v= (b_1 - b_2) u'_1 + b_2 u'_2 + ... + b_r u'_r$

Furthermore $gcd(b_1 - b_2, b_2, ..., b_r)=1$ and

$|b_1 - b_2| + |b_2| +...+ |b_r| \lt s$

so by inductive hypothesis the result follows.∎

Theorem: Let $F$ be a finitely generated free abelian group of rank $r$ and let $G$ be a subgroup of $F$ of rank $s$ with $0\lt s\le r$. Then there exist generators for $F$ $v_1,...,v_r$ such that

$G = \langle h_1 v_1 ,..., h_s v_s\rangle$

where $h_1,...,h_s$ are positive integers satisfying $h_i | h_{i+1}$ for $i = 1,...,s-1$.

Proof: Let $u_1,...,u_r$ be a set of generators for $F$. Take any $x\in G$. Write $x = x_1 u_1 +...+x_r u_r$. Define $\delta(x) = gcd(x_1,...,x_r)$. We claim that $\delta(x)$ is independent of the choice of generators of $F$.

This is easily seen because if $u'_1,...,u'_r$ are another set of generators, we can write the $u_i$'s in terms of the $u'_i$'s showing that $gcd(x_1,...x_r) | gcd(x'_1,...,x'_r)$ where $x = x'_1 u'_1 +...+x'_r u'_r$. By symmetry we must have equality.

Now take any nonzero $y_1\in G$ such that $\delta(y_1)$ is minimal. Set $h_1 = \delta(y_1)$. Then $y_1$ can be written $y_1 = h_1 (z_1 u_1 + ... + z_r u_r)$ for some integers $z_i$ satisfying $gcd(z_1,...,z_r) = 1$. By the lemma, there exist elements $v'_2,...v'_r$ which together with $v_1$ generate $F$.

Hence an element $y \in G$ can be written

$y = w_1 v_1 + w'_2 v'_2 +...+w'_r v'_r$

Now $h_1$ must divide $w_1$, since we have $w_1 = q h_1 + m$ for some $0 le 0 \lt h_1$ and $h_1$ is minimal. (Consider $\delta(y - q y_1)$.) Thus

$y - q y_1 = t_2 v'_2 + ... + t_2 v'_r$

If $r=1$ we are done, for we have $s=1, F=\langle v_1\rangle, G=\langle h_1 v_1\rangle$. We induct on $r$, so suppose $r \gt 1$.

Let $F_1 =\langle v_1 , v'_2 ,...,v'_r\rangle$ and $G_1 = F_1 \cap G$. Then $G_1$ is a subgroup of $F_1$ whose rank we shall denote by $t-1$ where $0 \lt t \le r$. If $t = 1$ then $G_1 = 0$ and since $G=\langle h v_1\rangle$ we are done. Otherwise $t\lt 1$, and by inductive hypothesis there exist free generators $v_2,...,v_r$ of $F_1$ such that

$G_1 = \langle h_2 v_2 ,...,h_t v_t \rangle$

where $h_i | h_{i+1}$ for $i=2,..,t-1$. Now $F = \langle v_1,...,v_r\rangle$ and any $y\in G$ can be written $y = q_1 h_1 v_1 + g_1$ for some $g_1 \in G_1$. Thus $h_1 v_1, ..., h_t v_t$ generate $G$. They must also be independent, becuause a nontrival relation between them imply a nontrivial relation between the generators $v_1,...,v_r$ of $F$.

Thus $G =\langle h_1 v_1,...,h_t v_t \rangle$ and $t = s$. It remains to show $h_1 | h_2$. Write $h_2 = a h_1 + b$ where $0 \le b \lt h_1$. Then consider $y_0 = h_1 v_1 + h_2 v_2 \in G$. We have $\delta(y_0) = gcd(h_1, h_2) = gcd(h_1, b)$. By minimality of $h_1$ we must have $b = 0$.∎