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Inverse image of a homomorphism

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  1. Jan 21, 2016 #1
    • Member warned about posting without the HW template
    The question: Let f: G -> H be a homomorphism of groups with ker(f) finite, the number of elements being n. Show that the inverse image is either empty or has exactly n elements.

    My work so far:
    Let h be eH (identity on H). Then the inverse image is ker(f) so has n elements, which makes it empty when n = 0.
    I know how to do this for eH but how for all other y in H?
     
  2. jcsd
  3. Jan 21, 2016 #2
    What makes you think any homomorphism is invertible?

    EDIT: Scratch that, I might have mis-understood.
    What is an inverse image, exactly? Do you mean the subset of H that is mapped from G?


    This is the kernel
    [itex]Ker f := \{g\in G | f(g) = e\in H\}[/itex]
    assuming I understand the problem correctly
    we also know that [itex]|Ker f| = n[/itex]

    The inverse image of just the unit element: [itex]f^{-1} (\{e\}) = \{e\in G, g_2,\ldots g_n\}\subset G[/itex]
     
    Last edited: Jan 21, 2016
  4. Jan 21, 2016 #3

    Samy_A

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    I assume you meant that the inverse image of an element of H is either empty or has exactly n elements.

    If ##f(x)=f(y)##, what is ##f(xy^{-1})##?
    What does the answer tell you about the set ##D=f^{-1}(h)##, where ##h \in Im(f)##?
     
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