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.

Quote:

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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.

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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 $\displaystyle G\twoheadrightarrow G/[G,G]$.

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.

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

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).

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Originally Posted by Deveno

why do we need to assume φ is onto?

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

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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.

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.

Quote:

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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.

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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?

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').