# Center of Symmetric Group

[SOLVED] Center of Symmetric Group

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

The attempt at a solution
It is obvious that e is in Z(Sn). If there is another element a ≠ e in Z(Sn), 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
Homework Helper
Can't you find two transpositions that don't commute?

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

Dick
Homework Helper
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?

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
Homework Helper
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.

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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
Homework Helper
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?

OK. Let g ≠ e be an arbitrary permutation in Sn, 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 Sn. 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 Sn, 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 Sn and so Z(Sn) = {e}.

How is that?

Dick