Commutator

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December 2nd 2011, 05:44 PM
dwsmith

Commutator

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

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

Correct?

And I need some help on this.
December 2nd 2011, 06:24 PM
Drexel28

Re: Commutator

Quote:

Originally Posted by dwsmith

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

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

Correct?

And I need some help on this.

You need to assume that $\phi$ is surjective. Then, your $\widehat{\phi}$ is surjective by mere definition! To factor $\widehat{\phi}$ through $G/[G,G]$ just do what's natural. What's the obvious map $G\twoheadrightarrow G/[G,G]$ .
December 3rd 2011, 12:32 AM
Deveno

Re: Commutator

why do we need to assume φ is onto?

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

the question is, is this mapping well-defined?

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

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

that is, G' ⊆ ker(φ).

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

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

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

nowhere did i use the surjectivity of φ.
December 3rd 2011, 08:17 AM
dwsmith

Re: Commutator

How do I show uniqueness. I tried assume another map pi but I wasn't doing ground breaking.
December 3rd 2011, 02:15 PM
Deveno

Re: Commutator

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

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

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

therefore:

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

( $\widehat{\phi}$ is unique because of the way we defined it, using the universal property of the canonical quotient map).
December 3rd 2011, 02:21 PM
Drexel28

Re: Commutator

Quote:

Originally Posted by Deveno

why do we need to assume φ is onto?

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

And this map would only be a surjection onto $\phi(G)$ which manes that this mapping is surjective if and only if $\phi$ is surjective. The universal property of the abelinization is stated for abelian images, for this reason.
December 3rd 2011, 02:34 PM
Deveno

Re: Commutator

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

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

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

$\widehat{\phi}(G') = (0,0), \widehat{\phi}((1 2)G') = (1,0)$ which is not surjective.
December 3rd 2011, 02:43 PM
Drexel28

Re: Commutator

Quote:

Originally Posted by Deveno

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

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

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

$\widehat{\phi}(G') = (0,0), \widehat{\phi}((1 2)G') = (1,0)$ 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 $G\to\phi(G)\hookrightarrow H$ ), but the operative thing, the way all the surjections work is if you are just dealing with abelian images. Agreed?
December 3rd 2011, 03:06 PM
Deveno

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').