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 $H$.

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

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?