Subset of the Group of Permutations: Subgroup or Not?

  • #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.

Homework Statement


[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]


Homework Equations


Not applicable.


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,
 
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Answers and Replies

  • #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.
 
  • #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.
 
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  • #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.
 
  • #5
StatusX
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Or just use restrict your example to the rational numbers.
 
  • #6
Tom Mattson
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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.
So it is! I am scraping the rust off slowly but surely...
 
  • #7
Tom Mattson
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As long as I've got the attention of you good people...

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


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.
 
  • #8
Dick
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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.
Sure. What could be wrong with that?
 

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