1. ## Commutator

Show that if $\displaystyle \phi:G\to H$ is a homomorphism into the abelian group H, then there exists a unique homomorphism $\displaystyle \hat{\phi}:G/G'\to H$ such that $\displaystyle \hat{\phi}(G'x)=\phi(x), \ \forall x\in G$.

I believe I need to show that phi is onto and there is a map $\displaystyle \theta:G\to G/G'$ such that $\displaystyle (\phi\theta)(g)=\hat{\phi}(G'x)$.

Correct?

And I need some help on this.

2. ## Re: Commutator

Originally Posted by dwsmith
Show that if $\displaystyle \phi:G\to H$ is a homomorphism into the abelian group H, then there exists a unique homomorphism $\displaystyle \hat{\phi}:G/G'\to H$ such that $\displaystyle \hat{\phi}(G'x)=\phi(x), \ \forall x\in G$.

I believe I need to show that phi is onto and there is a map $\displaystyle \theta:G\to G/G'$ such that $\displaystyle (\phi\theta)(g)=\hat{\phi}(G'x)$.

Correct?

And I need some help on this.
You need to assume that $\displaystyle \phi$ is surjective. Then, your $\displaystyle \widehat{\phi}$ is surjective by mere definition! To factor $\displaystyle \widehat{\phi}$ through $\displaystyle G/[G,G]$ just do what's natural. What's the obvious map $\displaystyle G\twoheadrightarrow G/[G,G]$.

3. ## Re: Commutator

why do we need to assume φ is onto?

i would simply define: $\displaystyle \widehat{\phi}(G'x) = \phi(x)$

the question is, is this mapping well-defined?

note that if $\displaystyle g \in G'$ then: $\displaystyle g = \prod_k a_kb_ka_k^{-1}b_k^{-1}$

hence $\displaystyle \phi(g) = \prod_k \phi(a_k)\phi(b_k)(\phi(a_k))^{-1}(\phi(b_k))^{-1} = e_H$, since H is abelian.

that is, G' ⊆ ker(φ).

now suppose that G'x = G'y, so that $\displaystyle xy^{-1} \in G'$

then $\displaystyle \phi(xy^{-1}) = e_H$ so $\displaystyle \phi(x)(\phi(y))^{-1} = e_H$ and therefore:

$\displaystyle \phi(x) = \phi(y)$, justifying our definition of $\displaystyle \widehat{\phi}$.

nowhere did i use the surjectivity of φ.

4. ## Re: Commutator

How do I show uniqueness. I tried assume another map pi but I wasn't doing ground breaking.

5. ## Re: Commutator

suppose $\displaystyle \psi$ is another such map, and let $\displaystyle \pi$ be the canonical projection $\displaystyle G \to G/G'$ given by $\displaystyle \pi(g) = G'g$.

then $\displaystyle \phi = \psi \circ \pi = \widehat{\phi} \circ \pi$.

since $\displaystyle \pi$ is surjective, it has a right-inverse, that is a function $\displaystyle \sigma$ such that $\displaystyle \pi \circ \sigma = \text{id}_{G/G'}$.

therefore:

$\displaystyle \psi \circ \pi \circ \sigma = \widehat{\phi} \circ \pi \circ \sigma$, so $\displaystyle \psi = \widehat{\phi}$

($\displaystyle \widehat{\phi}$ is unique because of the way we defined it, using the universal property of the canonical quotient map).

6. ## Re: Commutator

Originally Posted by Deveno
why do we need to assume φ is onto?

i would simply define: $\displaystyle \widehat{\phi}(G'x) = \phi(x)$
And this map would only be a surjection onto $\displaystyle \phi(G)$ which manes that this mapping is surjective if and only if $\displaystyle \phi$ is surjective. The universal property of the abelinization is stated for abelian images, for this reason.

7. ## Re: Commutator

but φ doesn't have to be onto. look, say i have this example:

$\displaystyle \phi:S_3 \to \mathbb{Z}_2 \times \mathbb{Z}_2$ given $\displaystyle \phi((1 2 3)) = (0,0), \phi((1 2)) = (1,0)$.

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 $\displaystyle \widehat{\phi}$ is given by:

$\displaystyle \widehat{\phi}(G') = (0,0), \widehat{\phi}((1 2)G') = (1,0)$ which is not surjective.

8. ## Re: Commutator

Originally Posted by Deveno
but φ doesn't have to be onto. look, say i have this example:

$\displaystyle \phi:S_3 \to \mathbb{Z}_2 \times \mathbb{Z}_2$ given $\displaystyle \phi((1 2 3)) = (0,0), \phi((1 2)) = (1,0)$.

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 $\displaystyle \widehat{\phi}$ is given by:

$\displaystyle \widehat{\phi}(G') = (0,0), \widehat{\phi}((1 2)G') = (1,0)$ which is not surjective.
I think we're saying the same thing. Of course you can make this map for any morphism (because you just go $\displaystyle G\to\phi(G)\hookrightarrow H$), but the operative thing, the way all the surjections work is if you are just dealing with abelian images. Agreed?

9. ## Re: 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').