The symmetric group on a set A is determined only by the cardinality of A for infinite sets A

Under the notation of the preceding two exercises (here and here), prove that |D| = |A| = |\Omega|. Deduce that if S_\Omega \cong S_\Delta then |\Omega| = |\Delta|. [Hint: Use the fact that D is generated by transpositions. You may assume that countable unions and finite direct products of sets of cardinality |\Omega| also have cardinality |\Omega|.]


Since D is generated by transpositions, every element of D is a finite product of transpositions. Thus |D| \leq |\bigcup_{n \in \mathbb{N}} \prod_{i=1}^n \Omega \times \Omega|. Because countable unions and finite direct products of sets of cardinality k have cardinality k, |D| \leq |\Omega|.

Now the set \Omega may be well ordered. For every element i \in \Omega, there is a permutation \sigma \in A such that i is minimal in M(\sigma) under this order; for example, choose any two elements j and k larger than i and let \sigma = (i\ j\ k). Thus |A| \geq |\Omega|.

Thus |\Omega| \leq |A| \leq |D| \leq |\Omega|, and we have |D| = |A| = |\Omega|.

Now let \Omega and \Delta be sets and suppose \varphi : S_\Omega \rightarrow S_\Delta is an isomorphism. We now prove a lemma.

Lemma: If \varphi : G \rightarrow H is a group isomorphism and N \leq G is a \leq-minimal nontrivial normal subgroup, then \varphi[N] \leq H is a \leq-minimal nontrivial normal subgroup. Proof: \varphi[N] is a subgroup since \varphi is a homomorphism and is normal since \varphi is surjective. Since |\varphi[N]| = |N|, \varphi[N] is nontrivial. Now suppose K \leq H is a nontrivial normal subgroup. \varphi^{-1} : H \rightarrow G is a group isomorphism, so that \varphi^{-1}[K] \leq G is a nontrivial normal subgroup, hence N \leq \varphi^{-1}[K]. Thus \varphi[N] \leq K, so that \varphi[N] \leq H is in fact a \leq-minimal nontrivial normal subgroup. \square

Now A_\Omega \leq S_\Omega is a minimal nontrivial normal subgroup, so that \varphi[A_\Omega] \leq S_\Delta is a minimal nontrivial normal subgroup. By this previous exercise, then, \varphi[A_\Omega] = A_\Delta. Thus |A_\Omega| = |A_\Delta|, and by the above argument, |\Omega| = |\Delta|.

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