Subset of the Group of Permutations: Subgroup or Not?

[quantumdude]

Well, in 5 years of PF'ing and watching over this forum, I am finally posting my first homework question. 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

S_A is the group of all permutations of a set A under permutation multiplication. B is a subset of A, and b is a particular element of B. Determine whether the given set is sure to be a subgroup of S_A under the induced operation. Here \sigma<b>=\{\sigma(x)|x \in B\}</b>

And the subsets are...

H=\{\sigma\in S_A|\sigma<b>\subseteq B\}</b>
K=\{\sigma\in S_A|\sigma<b>=B\}</b>

Homework Equations

Not applicable.

The Attempt at a Solution

First let's consider H. The elements of H are all of the permutations that send the elements of B to a subset of B. To try to grasp this, I considered an example.

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

\sigma_1=\left(\begin{array}{ccccc}1 &amp; 2 &amp; 3 &amp; 4 &amp; 5\\3 &amp; 1 &amp; 2 &amp; 5 &amp; 4 \end{array}\right)

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

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

I'll leave K alone until I get H sorted out.

Thanks,

 

[Dick]

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.

 

[quantumdude]

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 A=\mathbb{Z}, the set of integers. But your comment did get me thinking about other infinite sets, such as the reals.

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

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

That answers my question for K too, so if there are no objections to my reasoning from any math gurus then it looks like I can move on.

 

[Dick]

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.

 

[StatusX]

Or just use restrict your example to the rational numbers.

 

[quantumdude]

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...

 

[quantumdude]

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

K=\{\sigma\in S_A|\sigma<b>=B\}</b>

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

 

[Dick]

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



I would say that K is a subgroup of S_A here. Since the domain and the target are the same set (B), and since \sigma\in K is bijective, K 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?