MHB Properties of permutation of a set

[Kiwi1]

I am doing some self study of groups and can solve problem #3 but not Problem #4.

Problem 3.
Let A be a finite set, and B a subset of A. Let G be the subset of S_A consisting of all of the permutations f of A such that f(x) is in B for every x in B. Prove that G is a subgroup of S_A.

Problem 4.
Give an example to show that the conclusion of part 3 is not necessarily true if A is an infinite set.

Problem 3.
I can show that G is closed under composition and I know that inverses exist because f is a one-one permutation. So G is a sub group.

Problem 4.
First I can't see any implicit use of the fact that A is finite in my solution to part 3.

I believe that G has infinite members because for each member of G there are an infinite number of other members that permute the members of A-B in different ways.

If I let A be the positive rational no's and B =(0,1) then f(x)=x^n and x^(1/n) are in G. Seems OK.

 

[Evgeny.Makarov]

Problem 3.
Let A be a finite set, and B a subset of A. Let G be the subset of S_A consisting of all of the permutations f of A such that f(x) is in B for every x in B. Prove that G is a subgroup of S_A.

...

Problem 3.
I can show that G is closed under composition and I know that inverses exist because f is a one-one permutation.

You need to show not only that the inverse exists in $S_A$, but that it belongs to $G$.

 

[Kiwi1]

You need to show not only that the inverse exists in $S_A$, but that it belongs to $G$.

1. f is a bijection from A to A. So f inverse exists in A.
2. f is also injective from B to B because if it weren't then the image of B would have fewer elements than B (not certain if this is true if B has infinite elements).
3. f is also surjective from B to B because if it weren't then (being injective) the domain B would be larger than the image B (not certain if this is true if B has infinite elements).
4. So f is a bijective function from B to B. Hence it has an inverse in B.

So my argument relights upon counting the elements in B. Presumably this does not hold if A (and hence B) have an infinite number of elements.Q4. Let A be the integers and B the positive integers. One example of f is f = x^2. Now f^{-1} = x^{1/2} is not in G because sqrt(2) is not in G. So G is not closed under inverses when B is the positive integers.

Are my arguments sound?

 

[Evgeny.Makarov]

2. f is also injective from B to B because if it weren't then the image of B would have fewer elements than B (not certain if this is true if B has infinite elements).

Injectivity is always preserved when we take a restriction of a function.

3. f is also surjective from B to B because if it weren't then (being injective) the domain B would be larger than the image B (not certain if this is true if B has infinite elements).

Yes, here the finite size of $B$ is used.

Q4. Let A be the integers and B the positive integers. One example of f is f = x^2. Now f^{-1} = x^{1/2} is not in G because sqrt(2) is not in G. So G is not closed under inverses when B is the positive integers.

This $f$ is not a permutation of $A$.

 

[Kiwi1]

If I chose A to be the positive reals and B to be the positive integers then f will be a permutation in A and G will not be closed under inversion?

 

[Evgeny.Makarov]

Yes.