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Two Group homomorphism proofs

  1. Feb 5, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Let ##\theta : G \mapsto H## be a group homomorphism.
    A) Show that ##\theta## is injective ##\iff## ##\text{Ker}\theta = \left\{e\right\}##

    B) If ##\theta## is injective, show that ##G \cong Im \theta ≤ H##.

    3. The attempt at a solution

    A)The right implication is fine. Briefly, if ##\theta## is a homomorphism, then the identity in G is always sent to the identity in H. Since ##\theta## is injective, no other element maps there. I am not so sure about the left implication, but what I tried was:
    Consider ##\theta (g_1) = h_1## and ##\theta (g_2) = h_1. ## Want to show that ##g_1 = g_2## using the fact that ##\text{Ker}\theta = ##identity.
    Then ##\theta (g_1 e_G) = \theta (g_1) e_H = h_1 = \theta(g_2)e_H## so ##\theta(g_1) = \theta(g_2) (1)##. But I realise this is a trivial result. From here I want to show that ##g_1 = g_2##. We were given a hint to use inverses more explicitly, but I can't see it right now.

    B)I think I have this one, I just would like someone to check over it to make sure I haven't overlooked anything. So prove that ##G \cong Im \theta##

    This means ##G## is isomorphic to ##Im \theta## and so we have a bijective homomorphism. It is given that ##\theta## is injective, so we need only show surjectivity.
    But simply, by defintion of ##Im \theta##, it is surjective.

    Now prove ##Im \theta \leq H##, so use test for subgroup.
    i)Non empty: we are dealing with a homomorphism, so the identity in g is always mapped to identity in H, so the image is never nonempty.

    ii)Closure: Let ##h_1, h_2 \in Im \theta##. For some ##g_1, g_2 \in G, g_1 \neq g_2##, we have ##\theta(g_1) = h_1 ## and ##\theta(g_2) = h_2##. Then ##\theta(g_1)\theta(g_2) = h_1 h_2 = \theta(g_1 g_2). ## We found such an element that maps to ##h_1 h_2## so closure is satisfied.

    iii) Let ##\ell = h_1 h_2 \Rightarrow \ell h_2^{-1} = h_1 \Rightarrow \ell h_2^{-1} h_1^{-1} = e \Rightarrow \ell (h_1 h_2)^{-1} = e.## So we found an element such that when it is smashed with another element we get ##e \in Im \theta##, so closed under inverse.

    Is this okay? If so, it is really just the left implication of A) that I need a hint for.

    Many thanks
     
  2. jcsd
  3. Feb 5, 2013 #2
    For 1, what happens to the element [itex] g_1 g_2^{-1} [/itex] under the homomorphism?

    For 2, your proof is correct though you already know e is in the image as identity is preserved under the morphism. This is also a special case of the first isomorphism theorem.
     
  4. Feb 5, 2013 #3

    CAF123

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    Hi Kreizhn,

    So applying ##g_1 g_2^{-1}## to the homomorphism, I get $$\theta(g_1 g_2^{-1}) = \theta(g_1) \theta(g_2^{-1}) = \theta(g_1) (\theta (g_2))^{-1} = h_1 (h_1)^{-1} = e_H,$$ where I defined ##\theta(g_1) = \theta(g_2) = h_1##. Since we know only the identity in G, ##e_G## maps to ##e_H##, I conclude that ##g_1 = g_2##. Okay?
     
  5. Feb 5, 2013 #4
    Looks good.
     
  6. Feb 5, 2013 #5

    CAF123

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    Thanks for the help.
     
  7. Feb 14, 2013 #6

    CAF123

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    Apparently I had to show that there existed a homormorphism between G and Imθ.
     
  8. Feb 14, 2013 #7
    I don't follow: [itex] \theta [/itex] is your homomorphism, that is what you showed.
     
  9. Feb 15, 2013 #8

    CAF123

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    I don't think I showed ##\theta## was a homormorphism - it is given that ##θ## is a homormorphism. The comment on my work is that '##θ : G →H ## is a homormorphism, need to show homormorphism ##G →\operatorname{Im} \theta##'
     
  10. Feb 15, 2013 #9

    jbunniii

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    There's not much to show. We can certainly define ##\phi : G \rightarrow im(\theta)## by ##\phi(g) = \theta(g)##. Then ##\phi## is a homomorphism because ##\phi(ab) = \theta(ab) = \theta(a)\theta(b) = \phi(a)\phi(b)##. The only distinction between ##\phi## and ##\theta## is that ##\phi## is surjective even if ##\theta## is not. Note that ##\phi## is injective if and only if ##\theta## is.
     
  11. Feb 15, 2013 #10

    CAF123

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    Could you elaborate further how you made these deducements?
     
  12. Feb 15, 2013 #11

    jbunniii

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    OK. I defined ##\phi : G \rightarrow im(\theta)## by ##\phi(g) = \theta(g)##. If ##a \in im(\theta)## then by definition of the image, there is some ##g \in G## such that ##\theta(g) = a##. But this means ##\phi(g) = a##. This shows that ##\phi## is surjective.

    Suppose ##\theta## is injective. If ##\phi(g) = \phi(h)## then ##\theta(g) = \theta(h)##, so ##g = h## because of the injectivity of ##\theta##. This shows that ##\phi## is injective.

    Conversely, suppose ##\phi## is injective. If ##\theta(g) = \theta(h)##, then ##\phi(g) = \phi(h)##, and this implies ##g = h## because of the injectivity of ##\phi##. This shows that ##\theta## is injective.
     
  13. Feb 15, 2013 #12

    CAF123

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    Makes sense, thanks. But why did you define ##\phi(g) = \theta(g)##?Is this arbitrary?
     
  14. Feb 15, 2013 #13

    jbunniii

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    What other candidate is available if we want a homomorphism between ##G## and ##im(\theta)##? We are not given any information about ##G## or ##H##, other than the fact that ##\theta : G \rightarrow H## is a homomorphism. Therefore we have nothing to work with except ##\theta##. There may well be other homomorphisms from ##G## onto ##im(G)##, but no information is provided about any of them.
     
  15. Feb 15, 2013 #14
    I think the gist of it is this:

    Technically, when you restrict morphisms they could stop being morphisms. When you restrict the domain it may fail to be a subgroup so the map, while still defined, could fail to be a homomorphism. Conversely, if you restrict the codomain, the map may not have a well-defined target and so fail to be a mapping.

    In this instance, we are restricting the codomain so that [itex] \theta: G \to \text{im}\theta [/itex]. The only possible pitfall is that your target not be well-defined, for which it suffices to check that [itex] \text{im}\theta \subseteq \text{im}\theta[/itex]. But this is a tautology, so there's nothing to show.

    It seems incredibly pedantic, but perhaps if this is a "first course" in abstract algebra, the TA is just making sure you "dot your i's and cross your t's."
     
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