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Math Help - Homomorphism

  1. #1
    Mel
    Mel is offline
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    Homomorphism

    Totally lost again!
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    Last edited by Krizalid; January 29th 2009 at 10:17 AM.
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  2. #2
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    That's algebra!

    \forall a\in G_1,\forall (x,y)\in G,\ (x,y)(a,e_2)(x,y)^{-1}=(x,y)(a,e_2)(x^{-1},y^{-1}) =(xax^{-1},ye_2y^{-1})=(xax^{-1},e_2)
    a,x\in G_1 \Rightarrow xax^{-1}\in G_1 \Rightarrow (xax^{-1},e_2)\in N
    Therefore N is a normal subgroup of G.


    h:G_1\rightarrow N:x\mapsto (x,e_2) is surjective by definition of N , a group homomorphism because h(xy)=(xy,e_2)=(x,e_2)(y,e_2)=h(x)h(y) and injective because (x,e_2)=(e_1,e_2)\Rightarrow x=e_1 .
    We have found an isomorphism between G_1 and N.


    Consider a,b)\mapsto b" alt="\phi: G\rightarrow G_2a,b)\mapsto b" /> .
    \phi((a,b)(c,d))=\phi((ac,bd))=bd=\phi((a,b))\phi(  (c,d)) , so \phi is an homomorphism, which is surjective by definition of G.
    Moreover, \forall (a,b)\in G,\ \phi((a,b))=e_2\Leftrightarrow b=e_2 \Leftrightarrow (a,b)\in N.
    Hence \ker\phi = N , and we can conclude G/N\cong G_2 .
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