Help finding homomorphism

[Michael F. Stemper]

I am trying to find the homomorphism from S_3 to the subgroup
{(),(abc),(acb)}. Since it's normal in S_3, there must be
such a homomorphism. Call it "f".

Since o((abc))=o((acb))=3, we either have f((abc)) = (abc) or
f((abc)) = (acb).

What about the other elements? They're all transpositions, which
are involutions. Presumably, they must all map to involutions in
the image. But, there's only one involution in the image -- the
identity permutation.

This doesn't work, since if all of the transpositions were
in the kernel of the homomorphism (call it "f"), we would
have f((abc)) = f((ab)(bc)) = f((ab))f((bc)) = ()(), showing
that f maps all of S_3 to the identity.

There is obviously something very wrong with my understanding.
Would somebody please set me straight?


P.S. I do *not* want somebody to hand me the homomorphism. I just
want to have my mistake pointed out. If I don't work this out for
myself, I won't be any more knowledgeable than I am now.
--
Michael F. Stemper
There's no "me" in "team". There's no "us" in "team", either.

[Arturo Magidin]

On Monday, March 21, 2022 at 4:10:41 PM UTC-5, Michael F. Stemper wrote:
> I am trying to find the homomorphism from S_3 to the subgroup
> {(),(abc),(acb)}. Since it's normal in S_3, there must be
> such a homomorphism. Call it "f".

This assertion is false. What we know is that if G is a group and N is a normal subgroup, then there is a homomorphism from G to some group with KERNEL equal to N. there is no guarantee that there is a homomorphism to N.

The only normal subgroups of S_3 are the trivial group, S_3, and the subgroup { (), (abc), (acb)} that you describe. That means that all images of S_3 have order 1 (trivial map), 2 (if the kernel is this group) or 6. The only possible map from S_3 to a group of order 3 is, therefore, the trivial map.

> P.S. I do *not* want somebody to hand me the homomorphism. I just
> want to have my mistake pointed out.

Your mistake is the belief that such a map exists. The only homomorphism S_3 -> { (), (abc), (acb)} is the trivial map, sending everything to (). You have misunderstood the connection between normal subgroups and homomorphisms.

--
Arturo Magidin

[Michael F. Stemper]

Hey! Long time no see, and thanks for the reply.

A few hours of quality time with Dummit & Foote, Isaacs, and Pinter showed
me exactly what you said, although I kept trying to force it into my
mistaken understanding.

Ironically, I was doing this to try and improve my understanding
of quotient groups. I was right in needing to do that, but got off
on the wrong foot entirely.

Thanks again.

--
Michael F. Stemper
Psalm 94:3-6