What is the Center of the Symmetric Group when n ≥ 3?

[e(ho0n3]

[SOLVED] Center of Symmetric Group

Homework Statement
Show that for n ≥ 3, Z(S_n) = {e} where e is the identity element/permutation.

The attempt at a solution
It is obvious that e is in Z(S_n). If there is another element a ≠ e in Z(S_n), then... There must be some sort of contradiction and it has to do with the fact that n ≥ 3 but I can't figure it out. Any tips?

 

[Dick]

Can't you find two transpositions that don't commute?

 

[e(ho0n3]

I know that (12)(23) ≠ (23)(12). But how does this show that Z(S_n) is trivial?

 

[Dick]

You want to show that for any element g, except for e, there exists an element h such that gh!=hg. You've just done it if g is a transposition. I was thinking you could decompose g into disjoint cycles and construct an element that it doesn't commute with. It should work. Can you try it?

 

[e(ho0n3]

You want to show that for any element g, except for e, there exists an element h such that gh!=hg. You've just done it if g is a transposition.

I've only done it for a particular h and g. Not for every g.

I was thinking you could decompose g into disjoint cycles and construct an element that it doesn't commute with. It should work. Can you try it?

I'm thinking of the following: say the disjoint cycle decomposition of g is (ab...)(cd...). Let h = (ac). Then gh = (ad...) and hg = (ab...), which are not equal. Hmm...If this is to work in general, I will have to deal with the various ways in which g is decomposed.

 

[Dick]

You've done the case where g consists of a single transposition. Just generalize your point labelling. Now suppose g contains a cycle of three or more elements, say (123...). Hint: use only 1,2 and 3, that way your other permutation h will commute with all of the other cycles. That leaves only the case where g contains only transpositions, like (12)(34)... Can you find something that doesn't commute with that? Actually I think you are already on the right track. You don't need to know EVERYTHING about the cycle decomposition, you only need to know enough to i) find an h and ii) make sure your cases cover all permutations.

 

[e(ho0n3]

You've done the case where g consists of a single transposition. Just generalize your point labelling.

What do you mean by "point labelling"?

Now suppose g contains a cycle of three or more elements, say (123...). Hint: use only 1,2 and 3, that way your other permutation h will commute with all of the other cycles.

I have no idea what you mean by that hint. Anyways, if g has a cycle (abc...) then by letting h = (bc), h will not commute with g since gh(b) = x where x is not b or c and hg(b) = b. Is that what you meant?

That leaves only the case where g contains only transpositions, like (12)(34)... Can you find something that doesn't commute with that?

This is similar to what I wrote in my previous post. h = (13) will not commute with g = (12)(34)...

 

[Dick]

By 'point labelling', I just mean do what you are doing, call the elements a,b,c... instead of 1,2,3... Not that it really matters. I think you are doing all the right things. Can you just pull them altogether into a proof?

 

[e(ho0n3]

OK. Let g ≠ e be an arbitrary permutation in S_n, n ≥ 3, in disjoint cycle form. Now either (i) g has a 2-cycle, i.e. g = (ab)..., or (ii) g has a m-cycle where m ≥ 3, i.e. g = (abc...)...

Let h = (bc), c ≠ a. Obviously h is in S_n. Given (i), gh(b) = g(c) ≠ a since only g(b) = a, and hg(b) = h(a) = a. Given (ii) gh(b) = g(c) = x where x is neither b nor c (since only g(a) = b and g(b) = c) and hg(b) = h(c) = c. In either case, gh ≠ hg. Since g is arbitrary, then for every g in S_n, n ≥ 3, there is a permutation h such that gh ≠ hg. Ergo, there can't be a permutation other than e that commutes with all the permutations in S_n and so Z(S_n) = {e}.

How is that?

 

[Dick]

It looks fine. I might have stated it in a more awkward way with even more cases. But that's my problem.

 

[e(ho0n3]

Thanks a lot Dick.