## 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 . Deduce that if then . [Hint: Use the fact that is generated by transpositions. You may assume that countable unions and finite direct products of sets of cardinality also have cardinality .]

Since is generated by transpositions, every element of is a finite product of transpositions. Thus . Because countable unions and finite direct products of sets of cardinality have cardinality , .

Now the set may be well ordered. For every element , there is a permutation such that is minimal in under this order; for example, choose any two elements and larger than and let . Thus .

Thus , and we have .

Now let and be sets and suppose is an isomorphism. We now prove a lemma.

Lemma: If is a group isomorphism and is a -minimal nontrivial normal subgroup, then is a -minimal nontrivial normal subgroup. Proof: is a subgroup since is a homomorphism and is normal since is surjective. Since , is nontrivial. Now suppose is a nontrivial normal subgroup. is a group isomorphism, so that is a nontrivial normal subgroup, hence . Thus , so that is in fact a -minimal nontrivial normal subgroup.

Now is a minimal nontrivial normal subgroup, so that is a minimal nontrivial normal subgroup. By this previous exercise, then, . Thus , and by the above argument, .

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