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Math Help - Group Homomorphism

  1. #1
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    Group Homomorphism

    Hi anyone knows how to solve this? My answer was 3. Is it correct?

    Question:
    If f:S3-> Z2 given by
    f(a) = 1 if a is a 2-cycle
    f(a) = 0 otherwise is a group homomorphism.
    What is |ker(f)|?
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  2. #2
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    Re: Group Homomorphism

    Yes, it's 3. With such a small group, and with an explicit homomorphism given, you can explicitly see (and count) what the kernel is.

    But there's a more insightful way to see that the kernel has order 3. It's what you'll need when things aren't so simple.

    You should understand why this true:

    \text{Let } \phi : S_3 \rightarrow \mathbb{Z}_2 \text{ be *any* surjective homomorphism. Then } |ker(\phi)| = 3.

    You should also understand why this true:

    \text{Let } G_0, G_1 \text{ be *any* groups such that }|G_0| = 12,000,000, \  |G_1| = 1,000.

    \text{Let }\phi : G_0 \rightarrow G_1 \text{ be any surjective homomorphism. Then } |ker(\phi)| = 12,000.
    Last edited by johnsomeone; October 27th 2012 at 01:05 AM.
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