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Math Help - Number of different homomorphisms

  1. #1
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    Number of different homomorphisms

    How many homomorphisms exist from the abelian group Z_2\oplus Z_2 to S_3.

    My answer is - none.

    Since for every 1\neq a\in Z_2\oplus Z_2 we have a^2=1 then we must have f^2(a)=1 ( f is an homomorphism). so f(a) must be (12) or (23) or (13) but since a product of any of the above is not of order 2 then f would not be a homomorphism.

    Does it seem right or am I fundamentaly wrong?

    Any help appreciated
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  2. #2
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    Quote Originally Posted by skyking View Post
    How many homomorphisms exist from the abelian group Z_2\oplus Z_2 to S_3.

    My answer is - none.

    Since for every 1\neq a\in Z_2\oplus Z_2 we have a^2=1 then we must have f^2(a)=1 ( f is an homomorphism). so f(a) must be (12) or (23) or (13) but since a product of any of the above is not of order 2 then f would not be a homomorphism.

    Does it seem right or am I fundamentaly wrong?

    Any help appreciated


    You are wrong but I think not fundamentally: your argument about the order of the

    image is right, but you oversaw the fact that the order of any of the three 2-cycles

    you wrote there indeed is 2, so you can define f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23), and

    the same can be done with any of the other two elements of order 2 in that direct sum.

    Of course, the above is not the whole story since you still have

    to figure out how to define f on the other elements of that sum...

    Tonio
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  3. #3
    Senior Member Tinyboss's Avatar
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    Also, a "trivial" point: the zero homomorphism exists between any two groups.

    Edit: After rereading your post it seems that you got that already.
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  4. #4
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    Quote Originally Posted by tonio View Post
    You are wrong but I think not fundamentally: your argument about the order of the

    image is right, but you oversaw the fact that the order of any of the three 2-cycles

    you wrote there indeed is 2, so you can define f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23), and

    the same can be done with any of the other two elements of order 2 in that direct sum.

    Of course, the above is not the whole story since you still have

    to figure out how to define f on the other elements of that sum...

    Tonio
    OK, so this is what I get. Define Z_2\oplus Z_2 =\{(0,0),(0,1),(1,0),(1,1)\} =\{ e,a,b,c\}, so in order for f to be a homomorphism we must define f(e)=(1), f(a)=(mn), f(b)=(jk), f(c)=(rs) (where (mn),(jk),(rs) are 2-cycles in S_3). But the we get f(a+b)=f(c)\neq f(a)f(b) since (mn)(jk) is not a 2-cycle. similarly we can show the same for other combinations of a,b,c. Therefore f is not a homomorphism.

    So concluding we get that the only homomorphism is the trivial one.

    Am I missing something here?
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  5. #5
    Senior Member Tinyboss's Avatar
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    What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?
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  6. #6
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    Quote Originally Posted by Tinyboss View Post
    What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?
    It still would not be a homomorphism. for example if we define f(a)=(12) and everything else goes to (1), then we get (1)=f(c)=f(a+b)\neq (12)(1)=f(a)f(b).

    But if we choose 2 nonidentity elemnts from Z_2\oplus Z_2 and send them to the same 2-cycle (for example (12)) then we will get a homomorphism. For instance f(a)=f(b)=(12), then (1)=f(c)=f(a+b)=f(a)f(b)=(12)(12) and (12)=f(b)=f(a+c)=f(a)f(c)=(12)(1) and similarly for other combinations.

    So now in addition to the trivial homomorphism we get 3\times 3 choices for different homomorphisms.

    I think I am not missing anything now.

    Thanks
    Last edited by skyking; March 15th 2011 at 08:28 AM. Reason: typo
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