             Now suppose \(A\) to be partitioned into a (necessarily infinite) mutually disjoint set \(P\) of pairs. Take \(G\) to be the group of permutations of \(A\) which fix all the pairs in \(P\). Then \(P\in Sym(V)\); it can now be shown that \(Sym(V)\) contains no choice function on \(P\). For suppose \(f\) were a choice function on \(P\) and \(f \in Sym(V)\). Then \(f\) has a finite support which may be taken to be of the form \(\{a_{1}, \ldots, a_{n}, b_{1},\ldots,b_{n}\} with each pair \{a_{i}, b_{i}\} \in P\). Since \(P\) is infinite, we may select a pair \(\{c, d\} = U\) from \(P\) different from all the \(\{a_{i}, b_{i}\}\). Now we define \(\pi \in G\) so that \(\pi\) fixes each \(a_{i}\) and \(b_{i}\) and interchanges \(c\) and \(d\). Then \(\pi\) also fixes \(f\). Since \(f\) was supposed to be a choice function on \(P\), and \(U \in P\), we must have \(f(U) \in U\), that is, \(f(U) = c\) or \(f(U) = d\). Since \(\pi\) interchanges \(c\) and \(d\), it follows that \(\pi(f(U)) \ne f(U)\). But since \(\pi\) is an automorphism, it also preserves function application, so that \(\pi(f(U)) = \pi f(\pi(U))\). But \(\pi(U) = U\) and \(\pi f = f\), whence \(\pi(f(U)) = f(U)\). We have duly arrived at a contradiction, showing that the universe \(Sym(V)\) contains no choice function on \(P\).