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Thread: Number of different homomorphisms

  1. #1
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    Number of different homomorphisms

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

    My answer is - none.

    Since for every $\displaystyle 1\neq a\in Z_2\oplus Z_2$ we have $\displaystyle a^2=1$ then we must have $\displaystyle f^2(a)=1$ ($\displaystyle f$ is an homomorphism). so $\displaystyle f(a)$ must be $\displaystyle (12)$ or $\displaystyle (23)$ or $\displaystyle (13)$ but since a product of any of the above is not of order 2 then $\displaystyle f$ 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 $\displaystyle Z_2\oplus Z_2$ to $\displaystyle S_3$.

    My answer is - none.

    Since for every $\displaystyle 1\neq a\in Z_2\oplus Z_2$ we have $\displaystyle a^2=1$ then we must have $\displaystyle f^2(a)=1$ ($\displaystyle f$ is an homomorphism). so $\displaystyle f(a)$ must be $\displaystyle (12)$ or $\displaystyle (23)$ or $\displaystyle (13)$ but since a product of any of the above is not of order 2 then $\displaystyle f$ 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 $\displaystyle f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23)$, 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 $\displaystyle f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23)$, 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 $\displaystyle Z_2\oplus Z_2 =\{(0,0),(0,1),(1,0),(1,1)\} =\{ e,a,b,c\}$, so in order for $\displaystyle f$ to be a homomorphism we must define $\displaystyle f(e)=(1), f(a)=(mn), f(b)=(jk), f(c)=(rs)$ (where $\displaystyle (mn),(jk),(rs)$ are 2-cycles in $\displaystyle S_3$). But the we get $\displaystyle f(a+b)=f(c)\neq f(a)f(b)$ since $\displaystyle (mn)(jk)$ is not a 2-cycle. similarly we can show the same for other combinations of $\displaystyle a,b,c$. Therefore $\displaystyle f$ 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 $\displaystyle f(a)=(12)$ and everything else goes to $\displaystyle (1)$, then we get $\displaystyle (1)=f(c)=f(a+b)\neq (12)(1)=f(a)f(b)$.

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

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

    I think I am not missing anything now.

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