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**Deveno** basically he's saying any Sn for n > 2 has two elements that don't commute.

so let's do this:

given |S| > 2, we can put a copy of S3 inside A(S) by letting f do to s1,s2, and s3 what an element of S3 does to 1,2 and 3, and having f(x) = x for everything in S BUT s1,s2,s3.

so if we prove it for S3, we'll have proven it for all "bigger" permutation (symmetric) groups.

but (1 2)(1 3) = (1 2 3), while (1 3)(1 2) = (1 3 2), so (1 2) and (1 3) do not commute, done!

NOTE: i am using the convention herstein does for permutations, where ab means "first do a, then do b". be advised that many authors use the OPPOSITE convention,

where ab means first do b, then do a.