How many abelian groups of order 720 are there?
Let \(L, M, N\) be finitely generated \(\mathbb{Z}\)-modules, such that \(L \oplus M \cong L \oplus N\). Prove that \(M \cong N\).
Solution
Write \(C_m=\mathbb Z/m\mathbb Z\) as an additive group.
1. A finitely generated abelian group \(A\) has a decomposition\[ A\cong\mathbb Z^{r_A}\oplus \bigoplus_{p\ \mathrm{prime}}\ \bigoplus_{e\geq1} (C_{p^e})^{m_A(p,e)}, \]where \(r_A\) and all \(m_A(p,e)\) are nonnegative integers and onlyfinitely many multiplicities are nonzero. These integers areuniquely determined by \(A\). This is the elementary-divisor formof the structure theorem for finitely generated abelian groups[@DF04]. In particular, equality of all these invariants isequivalent to isomorphism.
2. There are exactly \(10\) abelian groups of order \(720\), up toisomorphism.
Since \(720=2^4\cdot3^2\cdot5\), finiteness forces \(r_A=0\), and the exponents of the cyclic factors for each prime must sum to that prime’s exponent in \(720\).
The partitions of \(4\) are exactly \[ (4),\quad(3,1),\quad(2,2),\quad(2,1,1),\quad(1,1,1,1). \] Indeed, a largest part of \(4\) or \(3\) leaves respectively \(0\) or \(1\); a largest part of \(2\) leaves either another \(2\) or two \(1\)’s; and otherwise every part is \(1\). They give the five possible \(2\)-primary groups \[ C_{16},\quad C_8\oplus C_2,\quad C_4\oplus C_4,\quad C_4\oplus C_2\oplus C_2,\quad C_2^{\oplus4}. \] The only partitions of \(2\) are \((2)\) and \((1,1)\), giving \(C_9\) and \(C_3\oplus C_3\) as the two possible \(3\)-primary groups. The \(5\)-primary group must be \(C_5\).
Taking one of the five groups in the first list, one of the two in the second, and their direct sum with \(C_5\) gives \(5\cdot2=10\) groups, each of order \(16\cdot9\cdot5=720\). Step 1 proves that this list is exhaustive and that distinct choices are nonisomorphic, since they give different prime-power multiplicities.
3. Finitely generated \(\mathbb Z\)-modules admit cancellationof a common direct summand.
Apply step 1 to \(L,M,N\). Combining their decompositions shows that direct sums add the free ranks and every prime-power multiplicity. Thus the isomorphism \(L\oplus M\cong L\oplus N\) and uniqueness give \[ r_L+r_M=r_L+r_N, \] and, for every prime \(p\) and positive integer \(e\), \[ m_L(p,e)+m_M(p,e)=m_L(p,e)+m_N(p,e). \] All these numbers are finite nonnegative integers. Subtracting the common terms gives \(r_M=r_N\) and \(m_M(p,e)=m_N(p,e)\) for all \(p,e\). Matching the summands in the two decompositions now gives \(M\cong N\). An additive-group isomorphism is \(\mathbb Z\)-linear because it preserves integer multiples, so this is the required module isomorphism.