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Guided proof to the isomorphism theorems.

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Let [itex]G_1[/itex] and [itex]G_2[/itex] be groups with normal subgroups [itex]H_1[/itex] and [itex]H_2[/itex], respectively. Further, we let [itex]\iota_1 : H_1 \rightarrow G_1[/itex] and [itex]\iota_2 : H_2 \rightarrow G_2[/itex] be the injection homomorphisms, and [itex]\nu_1 : G_1 \rightarrow G_1/H_1[/itex] and [itex]\nu_2 : G_2/H_2[/itex] be the quotient epimorphisms.

    Given that there exists a homomorphism [itex]\sigma : G_1 \rightarrow G_2[/itex], show that there exists a unique mapping [itex]\overline{\sigma} : G_1/H_1 \rightarrow G_2/H_2[/itex] such that [itex]\overline{\sigma} \circ \nu_1 = \nu_2 \circ \sigma[/itex] if and only if [itex]\sigma[H_1] \subset H_2[/itex]. If such a [itex]\overbar{\sigma}[/itex] exists, it is a homomorphism.


    2. Relevant equations

    There aren't any equations, as this is a proof.


    3. The attempt at a solution

    I know that since [itex]\nu_1[/itex] and [itex]\nu_2[/itex] are epimorphisms, they are surjective homomorphisms. So [itex]Im(\nu_1)=G_1/H_1[/itex] and [itex]Im(\nu_2)=G_2/H_2[/itex]. But I really don't see how to get this proof off the ground. Please help get me started.

    The next question reads as follows.

    Prove that there exists a unique mapping [itex]\sigma^{\prime} : H_1 \rightarrow H_2[/itex] such that [itex]\iota_2 \circ \sigma^{\prime} = \sigma \circ \iota_1[/itex] if and only if [itex]\sigma[H_1] \subset H_2[/itex]. If such a [itex]\sigma^{\prime}[/itex] exists, it is a homomorphism.
     
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2

    matt grime

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    Let s=sigma, v=v_1 and w=v_2, cos I want to do this without having to type in tex. The only place to start is with the composite ws, since that is a well defined map, and it goes from G_1 to G_2/H_2. This gives a map of G_1/H_1 if and only if H_1 is in the kernel of the map ws. Which is if and only if.... That is existence. Uniqueness we'll come to in a second.
     
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