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Math Help - Kernels and homomorphism of groups

  1. #1
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    Kernels and homomorphism of groups

    Let  (G,*) and  (H, \circ) be any two groups, and let  f:G x H \rightarrow \ G be given by  f(g,h) = g . Prove that f is a homomorphism of groups and give ker f.
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  2. #2
    Super Member Gamma's Avatar
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    we show it is a homomorphism.
    Let (g_1,h_1),(g_2,h_2) \in G \times H
    f((g_1,h_1)(g_2,h_2))=f((g_1 * g_2, h_1 \circ h_2))=g_1 * g_2 = f((g_1,h_1))*f((g_2,h_2))

    Ker(f) = \{(g,h)\in G \times H | f(g,h) = 0 \} = \{(g,h)\in G \times H | g=0 \}= \{0\} \times H
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  3. #3
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    Quote Originally Posted by funnyinga View Post
    Let  (G,*) and  (H, \circ) be any two groups, and let  f:G x H \rightarrow \ G be given by  f(g,h) = g . Prove that f is a homomorphism of groups and give ker f.
    let (a,b),(c,d) \in G \times H

    Then

     <br />
f[(a,b)\cdot(c,d)]=f[(a*c,b\circ d)]=a*c=f[(a,b)]*f[(c,d)]<br />

    So f is a homomorphism

    ker(f) is the set of all elements that get mapped to 0

    since the 2nd coordinate does not affect the immage of f

    ker(f)=\{ (0,b):b \in H\}
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  4. #4
    Super Member Gamma's Avatar
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    Generalization

    I was just thinking it might be of interest to you to note that this same concept can be generalized to multiple products and considering the mapping \pi_i : G_1 \times G_2 \times ... \times G_n \rightarrow G_i by \pi_i(g_1, g_2,..., g_i,..., g_n) = g_i

    This is a homomorphism and the kernel of this is G_1 \times G_2 \times ... \times \{0\} \times G_{i+1} \times ... \times G_n. And if you apply the first isomorphism theorem you know that \frac{G_1 \times G_2 \times ... \times G_n}{ker(\pi_i)} \cong Im(\pi_i) = G_i
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