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Math Help - Bijection and identity mappings

  1. #1
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    Bijection and identity mappings

    Let D be a set and suppose that \gamma : D \rightarrow D is a bijection which has the property that whenever  \delta : D \rightarrow D is a bijection, then \gamma \circ \delta = \delta \circ \gamma. If \gamma \neq Id_D, what can be said about |D|?

    I can't even think about where to start with this apart from that I think that  |D|\ge 2?
    Help would be much appreciated.
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    Quote Originally Posted by worc3247 View Post
    Let D be a set and suppose that \gamma : D \rightarrow D is a bijection which has the property that whenever  \delta : D \rightarrow D is a bijection, then \gamma \circ \delta = \delta \circ \gamma. If \gamma \neq Id_D, what can be said about |D|?

    I can't even think about where to start with this apart from that I think that  |D|\ge 2?
    Help would be much appreciated.

    If you know something about group theory and in particular about permutation groups, the above is equivalent to ask what

    permutation commutes with any other permutation, i.e. what is the center of the group S_X, where X can be any

    set. It is known, and fairly easy to show, that the center is always trivial if |D|>2 , so in your case we get that it

    must be |D| = 2 .

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by worc3247 View Post
    Let D be a set and suppose that \gamma : D \rightarrow D is a bijection which has the property that whenever  \delta : D \rightarrow D is a bijection, then \gamma \circ \delta = \delta \circ \gamma. If \gamma \neq Id_D, what can be said about |D|?

    I can't even think about where to start with this apart from that I think that  |D|\ge 2?
    Help would be much appreciated.
    To expand upon tonio's comment let #(D)\geqslant 3 and let \pi\in S_D=\left\{\pi\in D^D:\pi\text{ is a bijection}\right\} and

    \pi\ne\text{id}_D. We 'know' that \pi may be written as the product of disjoint cycles \pi=C_1C_2\cdots and by

    assumption that \pi\ne\text{id}_D we know that there exists some cycle C_k with \left|C_k\right|\geqslant 2. If \pi contains some

    C_k=(x,y) for some x,y\in D then it's easy to construct a non-commuting permutation, so assume that

    \pi does not contain a transposition (in its cycle decomposition). Then, by prior assumption we know that

    there exists some C_k with C_k=x\overset{\pi}{\mapsto}y\overset{\pi}{\mapsto}  z\overset{\pi}{\mapsto}\cdots with x,y,z distinct. Consider then the

    transposition \tau=(x,y). Note then that \pi(\tau(x))=\pi(y)=z but \tau(\pi(x))=\tau(y)=x from where

    it follows that \tau\pi\ne\pi\tau.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    If you know something about group theory and in particular about permutation groups, the above is equivalent to ask what

    permutation commutes with any other permutation, i.e. what is the center of the group S_X, where X can be any

    set. It is known, and fairly easy to show, that the center is always trivial if |D|>2 , so in your case we get that it

    must be |D| = 2 .

    Tonio
    Does this still hold if X is infinite (countable or uncountable)?
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    Quote Originally Posted by Swlabr View Post
    Does this still hold if X is infinite (countable or uncountable)?

    Good question: I think it does, but a combinatorial proof isn't available unless we limit ourselves to permutations

    moving a finite number of objects (because in that case we can work within S_n , for some natural n).

    So assume \pi\in S_X\,,\,\,|X|\geq \aleph_0\,,\,\,\pi(x)\neq x\,\,\forall x\in S , for at least an infinite countable set of elements S\subset X.

    But then we can find say x_1,x_2,x_3\in S s.t. \pi(x_i)\neq x_i\Longrightarrow \exists \sigma\in S_{X'_3}\,,\,\,X'_3:=\{x_i\,,\,\pi(x_i)\;;\;i=1,2,  3\}\,,\,s.t\,\,\sigma\overline{\pi}\neq \overline{\pi}\sigma , with \overline{\pi}:=\pi\mid_{S_{X'_3}}.

    As we can take \sigma as element of S_x we're then done.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    Good question: I think it does, but a combinatorial proof isn't available unless we limit ourselves to permutations

    moving a finite number of objects (because in that case we can work within S_n , for some natural n).

    So assume \pi\in S_X\,,\,\,|X|\geq \aleph_0\,,\,\,\pi(x)\neq x\,\,\forall x\in S , for at least an infinite countable set of elements S\subset X.

    But then we can find say x_1,x_2,x_3\in S s.t. \pi(x_i)\neq x_i\Longrightarrow \exists \sigma\in S_{X'_3}\,,\,\,X'_3:=\{x_i\,,\,\pi(x_i)\;;\;i=1,2,  3\}\,,\,s.t\,\,\sigma\overline{\pi}\neq \overline{\pi}\sigma , with \overline{\pi}:=\pi\mid_{S_{X'_3}}.

    As we can take \sigma as element of S_x we're then done.

    Tonio
    Why can't we just use the proof I used above? The only important part was that \#(D)\geqslant 3.
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    Quote Originally Posted by Drexel28 View Post
    Why can't we just use the proof I used above? The only important part was that \#(D)\geqslant 3.

    You used that the given permutation is a product of disjoing cycles. I can't see how can this be extended to, say and infinite

    number of cycles.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    You used that the given permutation is a product of disjoing cycles. I can't see how can this be extended to, say and infinite

    number of cycles.

    Tonio
    I agree that it's hard to make sense of disjoint if the set is infinite, but I still think the concept of partitioning up the set into orbits is still valid.
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    Quote Originally Posted by Drexel28 View Post
    I agree that it's hard to make sense of disjoint if the set is infinite, but I still think the concept of partitioning up the set into orbits is still valid.

    And I agree with that, yet I still can't see how your proof can be extended, with or without this, to the infinite case.

    Tonio
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