where \([j]_b\) denote the class of an integer \(j\) in \(\mathbb{Z}/b\text{.}\)
Proof.
Using Proposition 2.27, we let \(\psi: \mathbb{Z}\to \mathbb{Z}/(p_1^{e_1}) \times \cdots \times \mathbb{Z}/(p_l^{e_l})\) be the unique homomorhism that sends \(1\) to \(([1]_{{p_1}^{e_1}}, \cdots, [1]_{{p_l}^{e_l}})\text{.}\) Then
We see \(m \in \operatorname{ker}(\psi)\) and so \(\langle m \rangle \subseteq \operatorname{Ker}(\psi)\text{.}\) Conversely, if \(\psi(n) = 0\text{,}\) then \(p_i^{e_i} \mid n\) for all \(i\) and since \(p_1^{e_1}, \dots, p_l^{e_l}\) are pairwise relatively prime, it follows that \(m \mid n\text{.}\) This proves \(\operatorname{ker}(\psi) = \langle m \rangle\text{.}\) The claim follows by the First Isomorphism Theorem.
Remark7.25.
Sunzi’s Remainder Theorem frequently goes by the name of The Chinese Remainder Theorem, but it has been remarked that this is somewhat like referencing the Pythagorean Theorem as The Greek Triangle Theorem or Fermat’s Little Theorem as The French Power Postulate. In the interest of giving credit where credit is due, we include it here with the name of the 3rd-century Chinese mathematician, Sunzi, to whom the result is often attributed.
The following is the classification theorem for finitely generated abelian groups. We present it without proving it for now. The full proof will be given in the spring semester.
Theorem7.26.Fundamental Theorem of Finitely Generated Abelian Groups (FTFGAG).
Let \(G\) be a finitely generated abelian group. Then \(G\) is a direct product of cyclic groups. More precisely
There exist \(r,s \geq 0\text{,}\) prime integers \(p_1<\ldots< p_s\) and positive integers \(a_i \geq 1\) such that: \(G \cong \mathbb{Z}^r\times Q_1 \times \dots \times Q_s\) where \(|Q_i| = p_i^{a_i}\) for all \(i\text{.}\)
For each index \(1\leq i\leq s\text{,}\) there is a partition \(a_i = a_{i,1} + \dots + a_{i,j_i}\) with each \(a_{i,j} \geq 1\text{,}\) such that \(Q_i \cong (\mathbb{Z}/p_i^{a_{i1}}) \times \dots \times (\mathbb{Z}/p_i^{a_{i,j_i}})\text{,}\) thus overall we have
The \(r,p_i\)’s, \(j_i\)’s and \(a_{i,j}\)’s are uniquely determined by \(G\text{.}\)
equivalently, there exist \(r \geq 0, t \geq 0\text{,}\) and \(n_i \geq 2\) for all \(1\leq i\leq t\text{,}\) satisfying \(n_{i+1} \mid n_i\) for all \(i\) so that \(G \cong \mathbb{Z}^r \times (\mathbb{Z}/n_1) \times \dots \times (\mathbb{Z}/n_t)\text{.}\)
The integers \(r,s,n_1,\ldots ,n_t\) are uniquely determined by \(G\text{.}\)
Proof.
It suffices prove that for a given group \(G\text{,}\) we can recover its invariant factor form from its elementary divisor form, and vice versa. We will be a bit hand-wavey for this following the ideas from the above examples.
where \(d_1\) is the product of the elementary divisors of highest power for each distinct prime in the list \(p_1, \dots, p_l\text{,}\)\(d_2\) is the product of the next highest possible prime powers, and so on. We will have that \(d_2\mid d_1\) and in general that \(d_{i+1}\mid d_i\) since by definition the exponent of \(p_j\in d_i\) is greater or equal to the exponent of \(p_j\in d_{i+1}\text{.}\)
with \(d_1 \mid d_2 \mid \cdots \mid d_n\text{,}\) we may apply Sunzi’s Remainder Theorem to each \(\mathbb{Z}/d_i\) to find its elementary divisor form.
Example7.27.FTFGAG in Action.
For \(G\cong\mathbb{Z}/3\times \Z/5\times \Z/5\) we have \(Q_1=\mathbb{Z}/3\text{,}\)\(Q_2=\mathbb{Z}/5\times \Z/5\text{.}\)
In Theorem 7.26, the number \(r\) is the rank of \(G\text{,}\) the \(p_i^{a_{i,k}}\) are the elementary divisors of \(G\text{,}\) and the decomposition of \(G\) in parts (1-2) is called the elementary divisor decomposition of \(G\text{.}\) The decomposition in part (1) is also called a primary decomposition.
In Theorem 7.26, the number \(r\) is the rank of \(G\text{,}\) the numbers \(n_1,\ldots,n_t\) are the invariant factors of \(G\text{,}\) and the decomposition of \(G\) in part (1) is the invariant factor decomposition of \(G\text{.}\)
Example7.29.Elementary Divisor Form to Invariant Factor Form.
Since \(n_2:=(4 \cdot 9) \mid n_1:=(8 \cdot 27 \cdot 25)\text{,}\) this is in invariant factor form, and hence the rank of \(A\) is \(3\) and the invariant factors of \(A\) are \(4 \cdot 9\) and \(8 \cdot 27 \cdot 25\text{.}\)
Example7.30.Invariant Factor Form to Elementary Divisor Form.
We classify the abelian groups of order \(75\) up to isomorphism.
Let \(G\) be an abelian group of order \(75\text{.}\) Since \(G\) is finite the rank of \(G\) is \(r=0\text{.}\) Let’s determine the possible elementary divisors \(p_i^{a_{i,j}}\) so that
The above equation gives \(75=|G|= \prod_{k=1}^s\prod_{j=1}^{i_k}p_i^{a_{k,j}}\) and the possibilities for factoring \(75\) as a product of prime powers are \(75=3\cdot 5 \cdot 5\) or \(75=3\cdot 25\) which gives
Note that the two groups above are not isomorphic. To see this, note that there is an element of order \(25\) in \(\mathbb{Z}/25 \times \mathbb{Z}/3\text{,}\) namely \(([1]_{25},[0]_3)\) whereas every element \((a,b,c)\in \mathbb{Z}/5 \times \mathbb{Z}/5 \times \mathbb{Z}/3\) has order
