Non abelian groups of order 6 isomorphic to S_3

[alligatorman]

How can I show that all Non Abelian Groups of order 6 are isomorphic to S_3 without using Sylow's Theorems?

I have shown the following:
G has a non-normal group of normal subgroup of order 2
The elements of G look like:
1, a, a^2, b, c, d, where a,a^2 have order 3 and others have order 2

I want G to act on the set of left cosets of a subgroup of order 2 to get a homomorphism from G onto S3, but I'm kind of confused.

Any ideas? Thanks

 

[morphism]

G has a non-normal group of normal subgroup of order 2

Did you mean to say "G has a non-normal subgroup of order 2"?

I want G to act on the set of left cosets of a subgroup of order 2 to get a homomorphism from G onto S3, but I'm kind of confused.

That's a good idea. Call the subgroup G is acting on H. Look at the kernel of the induced homomorphism into S_3: it's a subgroup of H.

 

[alligatorman]

Oops, yes I meant non-normal s/gp of order 2.

So if H = (1,a) where a has order 2, and non-normal, and we let G act on the left cosets of H, then the kernel is $$\cap_{x\in G} xHx^{-1}$$, so the kernel is trivial.

I'm not sure where I'm going with this, which makes me think I did not fully understand your advice.

 

[morphism]

G has six elements. S_3 has six elements. We have an injective homomorphism from G into S_3. So ...!

 

[alligatorman]

That would imply isomorphism, but I'm confused as to why a trivial kernel implies that the homom. is injective. (assuming I did that correctly)

 

[morphism]

Try to prove it. It's very straightforward. Start with what it means for a homomorphism to be injective.

 

[alligatorman]

So suppose Ker p = 1 where p is homom.

Then let x,y be in G such that p(x)=p(y)

Then p(xy^-1)=p(x)p(y)^-1 =1 .. so x=y

So I guess this all proves that G is isomorphic to S3.

Thank you very much!