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Math Help - homomorphism from A4

  1. #1
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    homomorphism from A4

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    (1) there is no homomorphism from A4 onto a group of order 2, 4, or 6

    (2) there is a homomorphism from A4 onto a group of order 3.

    Can anyone please give me some hints ?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Let's see some work for the first one.

    Quote Originally Posted by xmcestmoi View Post

    (2) there is a homomorphism from A4 onto a group of order 3.

    Can anyone please give me some hints ?
    How much do you know about A_4? A (moderately) common fact is that A_4 has only one subgroup of order four (namely \{e, (12)(34), (13)(24), (14)(23)\}=N). But, it is fairly easy to prove that since N is the only subgroup of order four and conjugation by any element of that subgroup (i.e. gNg^{-1}) is another subgroup of order four that N must be invariant under conjugation. Thus, N\unlhd A_4. So, the canonical homomorphism is \theta:G\to G/N given by g\mapsto gN. This is a surjective homomorphism and |G/N|=[G:N]=\frac{12}{4}=3
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    Thank you !!
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  4. #4
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    Now I'm curious. How do you know there's no homorphism onto groups with order 4 or 6?
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  5. #5
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    Quote Originally Posted by maddas View Post
    Now I'm curious. How do you know there's no homorphism onto groups with order 4 or 6?
    The kernel of group homomorphism is a normal subgroup. As the previous reply said, there is only one normal subgroup of order 4 in A_4. If there were group homomorphisms onto groups with order 4 and 6, by the first isomorphism theorem, there should, at least, exist normal subgroups of order 3 and 2 in A_4. However, this is not the case here.
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