Let \(A,B,C\) be finite abelian groups. Prove that if \[ A\times B\cong A\times C, \] then \[ B\cong C. \]
Solution
1. Write the elementary-divisor decompositions\[ A\cong\prod_{p,k}(\ZZ/p^k\ZZ)^{a_{p,k}}, \qquad B\cong\prod_{p,k}(\ZZ/p^k\ZZ)^{b_{p,k}}, \qquad C\cong\prod_{p,k}(\ZZ/p^k\ZZ)^{c_{p,k}}, \]where the products range over primes \(p\) and integers \(k\geq1\), and all butfinitely many multiplicities are zero.
This is the elementary-divisor form of the fundamental theorem of finite abelian groups.
2. For every prime \(p\) and every \(k\geq1\),\[ a_{p,k}+b_{p,k} = a_{p,k}+c_{p,k}. \]
Taking direct products adds the multiplicities of each cyclic factor \(\ZZ/p^k\ZZ\). Hence \[ A\times B \cong \prod_{p,k}(\ZZ/p^k\ZZ)^{a_{p,k}+b_{p,k}}, \] while \[ A\times C \cong \prod_{p,k}(\ZZ/p^k\ZZ)^{a_{p,k}+c_{p,k}}. \] The assumed isomorphism \(A\times B\cong A\times C\) and uniqueness of the elementary-divisor decomposition force the displayed multiplicities to agree.
3. For every prime \(p\) and every \(k\geq1\),\[ b_{p,k}=c_{p,k}. \]
Cancel \(a_{p,k}\) from the equality in step 2.
4. One has\[ B\cong C. \]
By step 3, the elementary-divisor decompositions of \(B\) and \(C\) contain exactly the same cyclic \(p\)-power factors with the same multiplicities. Therefore the fundamental theorem of finite abelian groups gives \(B\cong C\).
5. Q.E.D.
Step 4 is the required conclusion.