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

[happyg1]

Homework Statement

Show that S_3 has the presentation &lt;x,y|x^3=y^2=(xy)^2=1&gt;

Homework Equations

x^{-1}=x^2,y^{-1}=y,xyxy=1
xyx=y^{-1}

The Attempt at a Solution

Let H=<x>, has at most order 3.
Then
y^{-1}xy=yxy=x^{-1}=x\in &lt; x &gt;
x^{-1}xx=x\in &lt; x &gt;

so
&lt;x&gt;\lhd G

Then let <y>=K
and use
If
H,K\subseteq G ,H\lhd G
then
G=&lt;x&gt;&lt;y&gt; \subseteq G
G=&lt;xy&gt;=&lt;x&gt;&lt;y&gt;
So
|G|\leq 6

Or I can write out all possible elements of the group
\{x,y,x^2,xy,x^2y,(xy)x^2\}
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 S_3 and G, but that seems a bit excessive.

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

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

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

y^{-1}xy=yxy=x^{-1}=x\in &lt; x &gt;

Has x referring to two different values. Splitting it into
y^{-1}xy=yxy=x^{-1}
and
x^{-1} \in &lt;x&gt;
is probably clearer.

 

[Hurkyl]

What do you know about kernels of homomorphisms?

 

[happyg1]

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

let N=&lt;ab|a^2=b^2=(ab)^2=1&gt;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 |G| \geq 6
putting that together with what I showed above, that |G| \leq 6 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

\{x,y,x^2,xy,x^2y,(xy)x^2\}

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.