Let \(n_1,\ldots,n_r\) be integers \(\geq2\). Prove that there is a finite group \(G\) containing elements \(g_1,\ldots,g_r\) such that \(g_i\) has exact order \(n_i\) for each \(i\), and \(g_ig_j\neq g_jg_i\) for \(i\neq j\).
Solution
Let \(T_1,\ldots,T_r\) be disjoint sets with \(\#T_i=n_i-1\). Let \(S\) be the union of the \(T_i\) together with one more element \(x\) outside all the \(T_i\). Let \(G\) be the set of permutations of \(S\).
Choose \(g_i\in G\) such that \(g_i\) acts as an \(n_i\)-cycle on \(T_i\cup\{x\}\), and acts as the identity on the complement. Then \(g_i\) has order \(n_i\). If \(i\neq j\), then \((g_ig_j)(x)=g_i(g_j(x))\in g_i(T_j)=T_j\), and similarly \((g_jg_i)(x)\in T_i\), so \(g_ig_j\neq g_jg_i\).