# Presentations of groups

1. Jun 25, 2007

### happyg1

1. The problem statement, all variables and given/known data

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

2. Relevant equations
$$x^{-1}=x^2,y^{-1}=y,xyxy=1$$
$$xyx=y^{-1}$$

3. The attempt at a solution

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

so
$$<x>\lhd G$$

Then let <y>=K
and use
If
$$H,K\subseteq G ,H\lhd G$$
then
$$G=<x><y> \subseteq G$$
$$G=<xy>=<x><y>$$
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

Last edited: Jun 25, 2007
2. Jun 25, 2007

### 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 $G<S_3$ 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.
Has $x$ referring to two different values. Splitting it into
$$y^{-1}xy=yxy=x^{-1}$$
and
$$x^{-1} \in <x>$$
is probably clearer.

3. Jun 25, 2007

### Hurkyl

Staff Emeritus
What do you know about kernels of homomorphisms?

4. Jun 29, 2007

### happyg1

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

let $$N=<ab|a^2=b^2=(ab)^2=1>$$

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.
thanks,
CC

Last edited: Jun 29, 2007
5. Jun 30, 2007

### 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.