Show that if is a homomorphism into the abelian group H, then there exists a unique homomorphism such that .

I believe I need to show that phi is onto and there is a map such that .

Correct?

And I need some help on this.

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- December 2nd 2011, 05:44 PMdwsmithCommutator
Show that if is a homomorphism into the abelian group H, then there exists a unique homomorphism such that .

I believe I need to show that phi is onto and there is a map such that .

Correct?

And I need some help on this. - December 2nd 2011, 06:24 PMDrexel28Re: Commutator
- December 3rd 2011, 12:32 AMDevenoRe: Commutator
why do we need to assume φ is onto?

i would simply**define**:

the question is, is this mapping well-defined?

note that if then:

hence , since H is abelian.

that is, G' ⊆ ker(φ).

now suppose that G'x = G'y, so that

then so and therefore:

, justifying our definition of .

nowhere did i use the surjectivity of φ. - December 3rd 2011, 08:17 AMdwsmithRe: Commutator
How do I show uniqueness. I tried assume another map pi but I wasn't doing ground breaking.

- December 3rd 2011, 02:15 PMDevenoRe: Commutator
suppose is another such map, and let be the canonical projection given by .

then .

since is surjective, it has a right-inverse, that is a function such that .

therefore:

, so

( is unique because of the way we defined it, using the universal property of the canonical quotient map). - December 3rd 2011, 02:21 PMDrexel28Re: Commutator
- December 3rd 2011, 02:34 PMDevenoRe: Commutator
but φ doesn't have to be onto. look, say i have this example:

given .

that is a perfectly good homomorphism, and it is NOT surjective.

now explicitly, G' = {e, (1 2 3), (1 3 2)}, as one can see by computing all 36 commutators. so in this case, we have G/G' = {G', (1 2)G'}.

so the map is given by:

which is not surjective. - December 3rd 2011, 02:43 PMDrexel28Re: Commutator
- December 3rd 2011, 03:06 PMDevenoRe: Commutator
the key fact is that the canonical projection is surjective. and that's all we care about, because we really only care about the homomorphic image of φ (whatever happens in the rest of H isn't of any interest to us, since it never gets "hit" with φ).

my point being, in actual practice, it might be easier to define φ and H, rather than define a priori some subgroup of H that φ maps surjectively to (we might only discover what this subgroup is by investigating φ in some detail).

in some problems in group theory, it is indeed key to require surjectivity (because you need to exhibit some pre-image). here, since "abelianness" is hereditary (for subgroups), we don't have to worry about it, we know the image of φ is abelian because it is a subgroup of an abelian group. so all we need to know is that φ factors through the abelianization (that is, that ker(φ) contains all of G').