Herstein: Simple proof with elements.

This is problem 15 on page 20 of Herstein's Abstract Algebra

If S has three or more elements, show that we can find f,g in A(S) such that fg doesn't equal gf.


Attempt:
So what this is saying that A(S) can't be a centralizer of S because fg doesn't = gf. And this group can't be abelian. And now I'm lost.

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.

Quote:

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Originally Posted by 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.

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This is a really great answer! Thank you.

I like the use of cycle notation.