Center of a symetric group

Iwasfrozen

Ok so here's the question:
For any group G, we define the center Z(G) to be
Z(g)={gϵG: gx=xg for all xϵG}

Let G=S_3. Prove that Z(s_3)={(1)}

Honestly I have no clue how to do this I know I need to apply the formula at the top but I don't know how. So far all I've done is found all six elements of S_3. Which are:

id
(1)(2,3)
(1,2)(3)
(1,2,3)
(1,3,2)
(1,3)(2).

But now I'm stuck. Any help will be appreciated.

hivizman

As there are only six elements in your set, it won't take long to work out the full 6x6 table applying the group operator. It may be helpful to rearrange the elements of the set as follows:

i = identity
a = (1)(2,3)
b = (2)(1,3)
c = (3)(1,2)
x = (1,2,3)
y = (1,3,2)

To make things simpler, you can make use of the fact that, for all j, k ε G, if jk ≠kj, then neither j nor k are in Z(G). Take any element other than the identity element (which must be in the center Z(G) by definition), for example a. Look at ab in comparison with ba, ac in comparison with ca, ax in comparison with xa and ay in comparison with ya. That should give you the required result.

Iwasfrozen

hivizman wrote: »

As there are only six elements in your set, it won't take long to work out the full 6x6 table applying the group operator. It may be helpful to rearrange the elements of the set as follows:

i = identity
a = (1)(2,3)
b = (2)(1,3)
c = (3)(1,2)
x = (1,2,3)
y = (1,3,2)

To make things simpler, you can make use of the fact that, for all j, k ε G, if jk ≠kj, then neither j nor k are in Z(G). Take any element other than the identity element (which must be in the center Z(G) by definition), for example a. Look at ab in comparison with ba, ac in comparison with ca, ax in comparison with xa and ay in comparison with ya. That should give you the required result.

Ok so I followed your advice above and this is what I got.

ab = (1,2,3) = x
ba = (1,3,2) = y So not equal

ac = (1,3,2) = y
ca = (1,2,3) = x So not equal

ax = (1,3)(2) = b
xa = (1,2)(3) = c So not equal

ay = (1,2)(3) = c
ya = (1,3)(2) = b So not equal

So since none of them are equal it means that a,b,c,x,y are not elements of Z(G)? Which means the only element of Z(G) is id. So Z(S_3) = {(id)} ? Is this right?

ZorbaTehZ

Note that if you can find an element that does not commute with all x \in G, then the centre is trivial. So have a look for such an element(s).

Iwasfrozen

ZorbaTehZ wrote: »

Note that if you can find an element that does not commute with all x \in G, then the centre is trivial. So have a look for such an element(s).

That's what I got, the only element in Z(S_3) is the id, so the center is trivial. Is that right?

hivizman

Iwasfrozen wrote: »

That's what I got, the only element in Z(S_3) is the id, so the center is trivial. Is that right?

Yes, that's right.

equivariant

ZorbaTehZ wrote: »

Note that if you can find an element that does not commute with all x \in G, then the centre is trivial. So have a look for such an element(s).

I think you meant "-- if you can not find an element that commutes with all x \in G, then the centre is trivial..."

hivizman

Both ZorbaTehZ and equivariant are correct.

The definition of a trivial center (one where the only element is the identity element) is where there is no element (other than the identity) in the group that commutes with all other elements in the group.

However, to demonstrate that the center is trivial, it is not necessary to check every element (other than the identity). So long as you can find any non-identity element that does not commute with all the other (non-identity) elements in the group, you can use the fact that I noted in post #2 (for all j, k ε G, if jk ≠kj, then neither j nor k are in Z(G)) to conclude that the center is trivial.

ray giraffe

hivizman wrote: »

Both ZorbaTehZ and equivariant are correct.

Both ZorbaTehZ and equivariant are slightly wrong.

ZorbaTehZ: You can't find an element that does not commute with all x in G, because every element commutes with the identity element.

You meant to say:
Note that if you can find an element that does not commute with all non-identity x in G, then the centre is trivial.

equivariant: Same reason. You meant to say "if you can not find a non-identity element that commutes with all x in G, then the centre is trivial"

equivariant

hivizman wrote: »

Both ZorbaTehZ and equivariant are correct.

The definition of a trivial center (one where the only element is the identity element) is where there is no element (other than the identity) in the group that commutes with all other elements in the group.

However, to demonstrate that the center is trivial, it is not necessary to check every element (other than the identity). So long as you can find any non-identity element that does not commute with all the other (non-identity) elements in the group, you can use the fact that I noted in post #2 (for all j, k ε G, if jk ≠kj, then neither j nor k are in Z(G)) to conclude that the center is trivial.

True. Perhaps I misinterpreted hivizman's post. I suppose that I would say " So long as you can find any non-identity element that does not commute with any other (non-identity) element in the group" rather than the above wording. I think that the wording above is a little misleading. Anyway, enough nitpicking from me
It is worth pointing out that having an element whose centraliser is trivial is stronger than having trivial center (I presume?), although I can't think of an example right now of a group with trivial center, all of whose elements have nontrivial centraliser. Anyone?

ZorbaTehZ

ray giraffe wrote: »

ZorbaTehZ: You can't find an element that does not commute with all x in G, because every element commutes with the identity element.

Technically no, because I said the centre would be trivial (i.e. it contains the identity element!) :pac: