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Order of image of homomorphism divides order of G

  1. Mar 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##\phi : G \rightarrow G'## be a group homomorphism. Show that if ##| G |## is finite, then ##| \phi [G] |## = ##| im (\phi)| ## is finite and a divisor of ##| G |##

    2. Relevant equations


    3. The attempt at a solution
    Assume that G is finite, with order n. Then im(G) is a subgroup of G' and has at most n elements, so must also be finite. Let im(G) have order m.

    Next, we know that ##\forall a \in G##, ##a^n = e##. Also, we know that ##\phi : G \rightarrow \phi [ G ]## is a surjection, so ##(\forall a' \in \phi [G]) (\exists a \in G)~ \phi (a) = a' ## Thus, ##a^n = e \Rightarrow \phi (a^n) = \phi (e) \Rightarrow \phi (a) ^n = e \Rightarrow (a') ^n = e##, where a' is any element of the image of ##\phi##. This means that n is a multiple of the order of a'. But m is also a multiple of the order of a'...

    This is as far as I get. I can't seem how to conclude that n is a multiple of m from this information
     
  2. jcsd
  3. Mar 5, 2017 #2

    fresh_42

    Staff: Mentor

    The trick is to consider ##G'/\Phi(G)##. This divides ##G'## into equivalence classes which are equally big.
     
  4. Mar 6, 2017 #3

    pasmith

    User Avatar
    Homework Helper

    If you could show a bijection from the collection of left cosets of [itex]\ker \phi[/itex] to [itex]\phi(G)[/itex], then Lagrange's Theorem would complete the proof.

    We are not given that [itex]G'[/itex] is finite.
     
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