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Math Help - Abelian factor group

  1. #1
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    Abelian factor group

    Prove that a factor group of an Abelian group is Abelian.

    I'm a bit confused by the whole factor group affair, now I assume G is an Abelian group, and suppose that H is a factor group of G.

    Then does that mean H is a normal subgroup of G such that G|H is a group?

    By a theorem, I know that G|H is a group under the operation (aH)(bH)=abH. Now, is this the right theorem for to use to solve this problem?

    Thanks.
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  2. #2
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    Let G be an abelian group, and let H be some subgroup.

    Every subgroup of an abelian group is normal, so H is normal. This means you can factor it out.

    Look at G/H and let p be the homomorphism that maps G to G/H.
    Basically what this means is that p(g1) = p(g2) only if g1 and g2 are in the same coset of H.

    Want to show that G/H is abelian.

    Take two elements in G/H. Why not p(g1) and p(g2)?
    Play with them to see if they commute. Recall with a homomorphism you can manipulate them as follows:

    p(g1)*p(g2) = p(g1*g2) = p(g2*g1) = p(g2)*p(g1)

    They do commute, so G/H is abelian.
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  3. #3
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    So a group is isomorphic to its factor group?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    So a group is isomorphic to its factor group?
    Not generally. But they are homomorphic, and I used this fact in the proof above.
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  5. #5
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    Or you can say the following.

    Let G be an abelian group and H a subgroup. Since all subgroups of abelian groups are normal it means G/H is a (well-defined) factor group. Let x_1,x_2 \in G/H meaning x_1 = (g_1H) and x_2 = (g_2H). Then x_1x_2 = (g_1H)(g_2H) = (g_1g_2)H = (g_2g_1)H = (g_2H)(g_1H)=x_2x_1.
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