Proving the Presentation of S_3 with <x,y|x^3=y^2=(xy)^2=1>

[happyg1]

Homework Statement

Show that [tex]S_3[/tex] has the presentation [tex]<x,y|x^3=y^2=(xy)^2=1>[/tex]

Homework Equations

[tex]x^{-1}=x^2,y^{-1}=y,xyxy=1[/tex]
[tex]xyx=y^{-1}[/tex]

The Attempt at a Solution

Let H=<x>, has at most order 3.
Then
[tex] y^{-1}xy=yxy=x^{-1}=x\in < x >[/tex]
[tex]x^{-1}xx=x\in < x > [/tex]

so
[tex]<x>\lhd G[/tex]

Then let <y>=K
and use
If
[tex] H,K\subseteq G ,H\lhd G[/tex]
then
[tex]G=<x><y> \subseteq G[/tex]
[tex]G=<xy>=<x><y>[/tex]
So
[tex]|G|\leq 6[/tex]

Or I can write out all possible elements of the group
[tex]\{x,y,x^2,xy,x^2y,(xy)x^2\}[/tex]
So the group presented has order of at most 6.(not sure if that's true)

My trouble comes when I try to show that it IS 6.

Do I list all of the cosets? How do I get equality so that I can show that this presentation is isomorphic to S3?

CC

 

[NateTG]

Well, if you're masochistic, you can write out the isomorphism and multiplication tables for [itex]S_3[/itex] and [itex]G[/itex], but that seems a bit excessive.

An alternative might be to show that there are only two possible groups of order [itex]6[/itex] and one of them is commutative.

Another option might be to show that [itex]G[/itex] and [itex]S_3[/itex] both have identical effective group actions on a set. Since you already know that [itex]|G|=6[/itex] showing that [itex]G<S_3[/itex] is sufficient, which, in turn can be showing by effective group action of [itex]G[/itex] on a set of order 3.

P.S.
Notation -- this is really not a big deal, but it bugs me.

[tex] y^{-1}xy=yxy=x^{-1}=x\in < x >[/tex]

Has [itex]x[/itex] referring to two different values. Splitting it into
[tex]y^{-1}xy=yxy=x^{-1}[/tex]
and
[tex]x^{-1} \in <x>[/tex]
is probably clearer.

 

[Hurkyl]

What do you know about kernels of homomorphisms?

 

[happyg1]

Ok here's what I came up with:
Consider:
[tex]a=(12) b=(123)[/tex]
then
[tex] a^2=(12)(12)=1;
b^3=(123)(123)(123)=1;
(ab)^2=(13)(13)=1[/tex]

let [tex]N=<ab|a^2=b^2=(ab)^2=1>[/tex]so then let x------>(12) and y------>(123)

F({xy})------->S3 is an onto homomorphism
and
G------->S3 is an onto homomorphism
but this implies that [tex]|G| \geq 6[/tex]
putting that together with what I showed above, that [tex]|G| \leq 6[/tex] this means that |G|=6 and the homomorphism from G to S3 is actually an isomorphism. (The kernel is contained in N)

I don't know for sure if my reasoning here is correct, and I'm not sure if it's completely clear.
Please give me some input.
thanks,
CC

 

[matt grime]

You say you wrote out all the elements

[tex]\{x,y,x^2,xy,x^2y,(xy)x^2\}[/tex]

in G. I don't see the identity in there...

The elements are:

e

then look at powers of x:

x,x^2

cos x has order 3

powers of y

y

cos y has order 2. Now that leaves mixed powers:

xy, x^2y

and we know that yx=x^2y, hence we can always write a word as one of the 6 listed items, e,x,x^2,y,xy,x^2y, and that these are all distinct. Clearly there are only 2 groups of order 6 (prove this - you simply need to look at what expressions xy might be eqaul to), this is one of them it isn't abelian, so it is S_3.