Important: Pick your favorite composite number \(m = \prod p_i^{e_i}\) and classify all abelian groups of that order.
- Write their invariant factor decompositions and their elementary divisor decompositions. Come up with an algorithm for converting back and forth between these.
Solution
Take \[ m=72=2^3\cdot3^2. \] By the classification theorem for finite abelian groups, the \(2\)-primary part is determined by a partition of \(3\), and the \(3\)-primary part by a partition of \(2\).
1. The possible primary components are\[ \begin{aligned} G_{(2)}&\cong C_8,\quad C_4\oplus C_2,\quad C_2\oplus C_2\oplus C_2,\\ G_{(3)}&\cong C_9,\quad C_3\oplus C_3. \end{aligned} \]
The partitions of \(3\) are \[ 3,\qquad2+1,\qquad1+1+1, \] which give the three \(2\)-primary groups. The partitions of \(2\) are \[ 2,\qquad1+1, \] which give the two \(3\)-primary groups.
2. Hence there are exactly six abelian groups of order \(72\).
The primary decomposition is unique, so each choice of a \(2\)-primary type and a \(3\)-primary type gives one isomorphism class, and every class occurs this way. Thus there are \[ 3\cdot2=6 \] classes.
3. Their elementary-divisor and invariant-factor forms are as follows:\[ \begin{array}{c|c} \text{elementary divisors}&\text{invariant factors}\\ \hline C_8\oplus C_9&C_{72}\\ C_8\oplus C_3\oplus C_3&C_3\oplus C_{24}\\ C_4\oplus C_2\oplus C_9&C_2\oplus C_{36}\\ C_4\oplus C_2\oplus C_3\oplus C_3&C_6\oplus C_{12}\\ C_2\oplus C_2\oplus C_2\oplus C_9&C_2\oplus C_2\oplus C_{18}\\ C_2\oplus C_2\oplus C_2\oplus C_3\oplus C_3&C_2\oplus C_6\oplus C_6 \end{array} \]
For relatively prime integers \(a,b\), one has \[ C_a\oplus C_b\cong C_{ab}. \] To obtain invariant factors, align the prime-power cyclic factors in nondecreasing order, pad the shorter lists on the left by \(1\)’s, and multiply columnwise. For example, \[ (C_4,C_2)=(C_2,C_4), \qquad (C_9)=(C_1,C_9), \] so \[ (C_2\oplus C_4)\oplus C_9 \cong C_{2\cdot1}\oplus C_{4\cdot9} =C_2\oplus C_{36}. \] The other rows follow identically. In every invariant-factor row, each factor divides the next.
4. The conversion algorithm from elementary divisors to invariant factors is columnwise multiplication of aligned prime powers.
For each prime \(p\), write the \(p\)-primary cyclic factors as \[ p^{a_{p,1}}\mid p^{a_{p,2}}\mid\cdots\mid p^{a_{p,r_p}}. \] Let \(r=\max_p r_p\) and pad each list on the left by exponent \(0\). For each column \(j\), define \[ d_j=\prod_p p^{a_{p,j}}. \] Then \[ d_1\mid d_2\mid\cdots\mid d_r, \] and the direct sum of the \(C_{d_j}\) is the invariant-factor decomposition.
5. The reverse algorithm factors each invariant factor into its prime-power parts.
Given \[ C_{d_1}\oplus\cdots\oplus C_{d_r}, \qquad d_1\mid\cdots\mid d_r, \] factor each \[ d_j=\prod_p p^{v_p(d_j)}. \] Using the Chinese remainder theorem, \[ C_{d_j}\cong\bigoplus_{p\mid d_j}C_{p^{v_p(d_j)}}. \] Collecting these prime-power cyclic summands gives the elementary-divisor decomposition. The two algorithms are inverse because they merely regroup the same primary cyclic factors.