Results 1 to 9 of 9

Thread: Commutator

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: Commutator

    Quote Originally Posted by dwsmith View Post
    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 $\displaystyle G\twoheadrightarrow G/[G,G]$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Re: Commutator

    How do I show uniqueness. I tried assume another map pi but I wasn't doing ground breaking.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: Commutator

    Quote Originally Posted by Deveno View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: Commutator

    Quote Originally Posted by Deveno View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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').
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. commutator ring
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Aug 17th 2011, 02:07 PM
  2. Lie derivative and commutator
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 23rd 2009, 01:08 AM
  3. Commutator, Abelian
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Jun 2nd 2009, 11:18 PM
  4. commutator subgroup eep!
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Apr 30th 2009, 08:02 PM
  5. commutator subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 14th 2008, 10:40 AM

/mathhelpforum @mathhelpforum