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Subset of the Group of Permutations: Subgroup or Not?

  1. Feb 27, 2007 #1

    Tom Mattson

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    Well, in 5 years of PF'ing and watching over this forum, I am finally posting my first homework question. :tongue2: I'm taking a graduate course in Algebra, and it's been 11 years since I took the undergraduate version. So, I'm going back and doing all the homework exercises in my undergrad book. I'm stuck on this one.

    1. The problem statement, all variables and given/known data
    [itex]S_A[/itex] is the group of all permutations of a set [itex]A[/itex] under permutation multiplication. [itex]B[/itex] is a subset of [itex]A[/itex], and [itex]b[/itex] is a particular element of [itex]B[/itex]. Determine whether the given set is sure to be a subgroup of [itex]S_A[/itex] under the induced operation. Here [itex]\sigma=\{\sigma(x)|x \in B\}[/itex]

    And the subsets are...

    [itex]H=\{\sigma\in S_A|\sigma\subseteq B\}[/itex]
    [itex]K=\{\sigma\in S_A|\sigma=B\}[/itex]


    2. Relevant equations
    Not applicable.


    3. The attempt at a solution
    First let's consider [itex]H[/itex]. The elements of [itex]H[/itex] are all of the permutations that send the elements of [itex]B[/itex] to a subset of [itex]B[/itex]. To try to grasp this, I considered an example.

    Let [itex]A=\{1,2,3,4,5\}[/itex] and [itex]B=\{1,2,3\}[/itex]. Then choose a permutation [itex]\sigma_1[/itex] that satisfies the condition of membership in [itex]H[/itex].

    [tex]\sigma_1=\left(\begin{array}{ccccc}1 & 2 & 3 & 4 & 5\\3 & 1 & 2 & 5 & 4 \end{array}\right)[/tex]

    When I look at this, I can't see how [itex]H[/itex] could be anything other than [itex]K[/itex] itself. If the image of [itex]B[/itex] under [itex]\sigma_1[/itex] is anything other than [itex]B[/itex], then it contains elements of [itex]A[/itex] that do not belong to [itex]B[/itex]. Hence, the image would not be a subset of [itex]B[/itex].

    The answer in the back of the book says that [itex]H[/itex] is not a subgroup of [itex]S_A[/itex], as it is not closed under taking of inverses. I do not see how that could possibly be right.

    I'll leave [itex]K[/itex] alone until I get [itex]H[/itex] sorted out.

    Thanks,
     
    Last edited: Feb 27, 2007
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  3. Feb 27, 2007 #2

    Dick

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    Hi Tom! Congratulations on your first post! I think your problem is that you are assuming that A and B are finite. They don't have to be.
     
  4. Feb 27, 2007 #3

    Tom Mattson

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    Eureka! I've got it!

    Dick, it's true that I was assuming that the set was finite. But that wasn't my problem. I would get the same result if I had let [itex]A=\mathbb{Z}[/itex], the set of integers. But your comment did get me thinking about other infinite sets, such as the reals.

    If I let [itex]A=\mathbb{R}[/itex], [itex] B=[0,1] [/itex], and [itex]\sigma(x)=\frac{1}{2}x[/itex], then the image of [itex]B[/itex] under [itex]\sigma[/itex] is [itex][0,\frac{1}{2}]\subset B[/itex]. But the [itex]\sigma^{-1}(x)=2x[/itex] does not map [itex]B[/itex] onto a subset of itself.

    My problem was in fact that I was assuming that the set [itex]A[/itex] is countable. But your comment did knock my noggin loose from being stuck in finite countable sets, so thanks for that. :approve:

    That answers my question for [itex]K[/itex] too, so if there are no objections to my reasoning from any math gurus then it looks like I can move on.
     
    Last edited: Feb 27, 2007
  5. Feb 27, 2007 #4

    Dick

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    It's not a countability issue. Take A to be the integers and B to be the even integers. Then take sigma(k)=2*k on B. Define sigma for the odd integers any which way, so its a bijection with Z-2*B (you know it's possible and the details aren't important). Same problem.
     
  6. Feb 27, 2007 #5

    StatusX

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    Or just use restrict your example to the rational numbers.
     
  7. Feb 27, 2007 #6

    Tom Mattson

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    So it is! I am scraping the rust off slowly but surely...
     
  8. Feb 27, 2007 #7

    Tom Mattson

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    As long as I've got the attention of you good people...



    I would say that [itex]K[/itex] is a subgroup of [itex]S_A[/itex] here. Since the domain and the target are the same set ([itex]B[/itex]), and since [itex]\sigma\in K[/itex] is bijective, [itex]K[/itex] is closed under taking of inverses. And of course, it satisfies the other conditions for being a subgroup as well.
     
  9. Feb 27, 2007 #8

    Dick

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    Sure. What could be wrong with that?
     
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