[FOM] Universe-sized groups with only set-sized normal subgroups
David Diamondstone
ddiamondstone at gmail.com
Sun Nov 20 18:20:26 EST 2011
> Let k be an inaccessible cardinal, G a group with |G| ≥ k. Say G is k-simple if for all normal N < G, |N| < k. Can we construct a k-simple group? Can we force this?
>
There are arbitrarily large simple groups, so yes (in an uninteresting way). You can construct examples by generalizing the alternating groups.
Let X be any set. We say that a permutation of X is even if it is the product of an even number of transpositions (so even permutations of X can only move finitely many elements around). Let A_X be the set of all even permutations of X. Since elements of A_X can be put in finite-to-one correspondence with finite subsets of X, the cardinality of A_X equals the cardinality of X (assuming X is infinite). Furthermore A_X is naturally a group under composition.
Claim: A_X is simple.
Proof: Let p,q be non-identity elements of A_X. There is some finite subset F of X such that p,q are the identity outside of F, and |F|≥5. Let A_F be the subgroup of A_X consisting of permutations which are the identity outside of F. A_F is naturally isomorphic to the alternating group A_n, where n=|F|. By the simplicity of A_n, q can be written as a product of conjugates of p and p^-1, hence any normal subgroup containing p must contain q.
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