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Math Help - Group Homomorphisms

  1. #1
    Ife
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    Group Homomorphisms

    Let \varphi :G\rightarrow H be a group homomorphism. Show that \varphi[G] is abelian if and only if for all x,y\in G, we have xyx^{-1}y^{-1}\in Ker\left ( \varphi \right )
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Group Homomorphisms

    let [tex]\varphi (G) [tex] be abelian want for [tex]x,y \in G [tex] we have xyx^{-1}y^{-1}\in Ker(\varphi)
    you should know that
    Ker(\varphi) = \{ g \in G \mid \varphi(g) = e_H\}
    so if we prove that \varphi (xyx^{-1}y^{-1} ) = e_H
    we are finished \varphi is homo which means \varphi (ab)= \varphi (a) \varphi(b) and \varphi(a^{-1}) = ( \varphi(a))^{-1}

    not the converse direction for all  x,y \in G we have xyx^{-1}y^{-1} \in Ker(\varphi)
    want \varphi(G) abelian
    let a,b \in \varphi (G) \Rightarrow \varphi^{-1} (a) , \varphi^{-1}(b)  \in G

    \varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1} \in Ker(\varphi)

    \varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1}] = e_H

    but it is homo

    \varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a^{-1})) (\varphi^{-1}(b^{-1}))] = e_H

    aba^{-1}b^{-1} = e_H \Rightarrow ab = ba
    the proof ends
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  3. #3
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    Re: Group Homomorphisms

    this is just the same thing as Amer said:

    suppose that \varphi(G) is abelian. then for any x,y in G:

    \varphi(xyx^{-1}y^{-1}) = \varphi(x)\varphi(y)\varphi(x^{-1})\varphi(y^{-1}) (since φ is a homomorphism)

    =\varphi(x)\varphi(y)(\varphi(x))^{-1}(\varphi(y))^{-1} (again, since φ is a homomorphism)

    = \varphi(x)(\varphi(x))^{-1}\varphi(y)(\varphi(y))^{-1} (since φ(G) is abelian).

     = e_He_H = e_H showing that xyx^{-1}y^{-1} \in ker(\varphi)

    on the other hand suppose that xyx^{-1}y^{-1} \in ker(\varphi) for all x,y in G.

    for any g,h in φ(G), we have φ(x) = g and φ(y) = h, for SOME x,y in G (because φ is onto φ(G)). but we know that xyx^{-1}y^{-1} \in ker(\varphi),

    so \varphi(xyx^{-1}y^{-1}) = e_H. but:

    \varphi(xyx^{-1}y^{-1}) = \varphi(x)\varphi(y)(\varphi(x))^{-1}(\varphi(y))^{-1}

    = ghg^{-1}h^{-1} = e_H, which quickly gives:

    gh = hg

    the quantity xyx^{-1}y^{-1} is called the commutator of x and y, and is written [x,y]. and here's a sneak peek of what lies ahead:

    the subgroup of G generated by all the commutators, is called the commutator subgroup [G,G] (or sometimes G'). and what you (we?) have just proved is that
    G/[G,G] is an abelian group (think about this for a second until it dawns on you what i am saying is true).

    well, we've almost proved that. we haven't yet shown that [G,G] is normal in G, but it is, and if φ:G-->H is a homomorphism, and H is abelian, then ker(φ) contains [G,G]. [G,G] measures how "non-abelian" G is, in a natural way (the center of G,Z(G), does the same thing, but in a way that is "not natural").

    if we write a^g for gag^{-1}, try proving that:

    [a,b]^g = [a^g,b^g]. this shows [G,G] is normal (why?).
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