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Homomorphisms and Calculating Elements

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that ##G## is a cyclic group with generator ##g##, that ##H## is some arbitrary group, and that ##\phi : G \rightarrow H## is a homomorphism. Show that knowing ##\phi (g)### lets you compute ##\phi(g_1)## ##\forall g_1 \in G##

    2. Relevant equations

    ##\phi(g^n) = \phi(\underbrace{g~\star_G~ g~ \star_G~ g ... \star_G~ g}_{n|}) = \underbrace{\phi (g) \star_H \phi(g) \star_H ... \star_H \phi(g)}_{|n|} = \phi(g)^n ##

    3. The attempt at a solution
    Let ##g_1 \in G## be arbitrary. This implies that ##\exists k \in \mathbb{Z}## such that ##g_1 = g^k##.

    ##\phi(g_1 \star_G g ) = \phi(g_1) \star_H \phi(g) \iff##

    ##\phi(g^k \star_G g) = \phi(g_1) \star_H \phi(g) \iff##

    ##\phi(g^{k+1}) = \phi(g_1) \star_H \phi(g) \iff##

    ##\phi(g)^{k+1} \star_H \phi(g)^{-1} = \phi(g_1) \iff##

    ##\phi(g)^k = \phi(g_1)##

    This seems correct, but I am unsure. If this is correct, though, does this imply that ##H## is also cyclic, where ##\phi(g)## is the generator?
     
  2. jcsd
  3. Nov 13, 2014 #2

    Dick

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    Science Advisor
    Homework Helper

    Wouldn't it be much simpler just to say ##\phi(g_1)=\phi(g^k)=\phi(g)^k##?? I don't know what all the extra stuff is for. And no, it doesn't mean H is cyclic. It does mean that the image of G under ##\phi## is a cyclic subgroup of H.
     
  4. Nov 14, 2014 #3
    Hmm...yes, it would have been terribly simpler, had I done that...
     
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