Show that if \(G = G_1 \oplus G_2\), where \(G_1\) and \(G_2\) are cyclic of orders \(m\) and \(n\), respectively, then \(m\) and \(n\) are not uniquely determined by \(G\) in general. [Hint: If \(m\) and \(n\) are relatively prime, show that \(G\) is cyclic of order \(mn\).]
Solution
Let \[ G_1\cong C_m,\qquad G_2\cong C_n, \] with \(\gcd(m,n)=1\). Then the element \[ (\bar 1,\bar 1)\in C_m\oplus C_n \] has order \(\operatorname{lcm}(m,n)=mn\). Since \(|C_m\oplus C_n|=mn\), this element generates the whole group. Hence \[ C_m\oplus C_n\cong C_{mn}. \] Equivalently, this is the Chinese remainder isomorphism \[ \mathbb Z/(mn)\longrightarrow \mathbb Z/m\oplus\mathbb Z/n, \qquad [a]_{mn}\longmapsto([a]_m,[a]_n). \]
Therefore the orders of two cyclic direct-summand factors are not determined by the abstract group. For example, \[ C_2\oplus C_{15}\cong C_{30}\cong C_3\oplus C_{10}, \] while the unordered pairs of factor orders are \(\{2,15\}\) and \(\{3,10\}\), which are different.
Thus \(m\) and \(n\) are not uniquely determined by \(G\) in general.