Here's the setup. I'm done reading about functions and permutations. In the homework problems for permutations, I'm on a section that focuses on the properties of permutations of a set A. The symmetric group on A, S

_{A}, is the set of all the permutations of A. In all the problems, G is a subset of S

_{A}, and you have to prove that G is a subgroup of S

_{A}given specific information. I'm stuck on the 4th problem. I'll go through #1-3 first, giving informal proofs along the way.

1. Let A be a set and a be an element of A. Let G be the subset of S

_{A}consisting of all the permutations f of A such that f(a)=a. Prove that G is a subgroup of S

_{A}. To prove this, you have to show closure for operation (composition) and for inverses.

- Let g be an element of G. Then g(a)=a. f(g(a))=f(a)=a, and g(f(a))=g(a)=a. Therefore, G is closed with respect to composition.
- If f(a)=a, then f
^{-1}(a)=a. Therefore, G is closed under inverses.

*f moves a*if f(a) doesn't equal a. Let A be an infinite set, and let G be the subset of S

_{A}which consists of all the permutations f of A which move

*only a finite number of elements*of A. Prove that G is a subgroup of S

_{A}.

- When performing fog and gof, the number of elements that get moved are
*at most*the combined total number of finite moveable elements of each permutation. So if {a_{1},..., a_{m}} are the moveable elements of f and {b_{1},...,b_{n}} are the moveable elements of g, and both sets of moveable elements are mutually exclusive; then fog and gof have at most the combined total number from both sets of finite moveable elements. Since this number of elements is finite, composition is closed for G. - Elements of f and g that don't move are of the form f(x)=x. This keeps the images of the moveable elements within the set of finite moveable elements because f
^{-1}(x)=x and permutations are bijective. Therefore, if a_{i}and a_{j}are moveable elements of f, then f(a_{i})=a_{j}and f^{-1}(a_{j})=a_{i}. The inverse of a f has a finite number of moveable elements. Hence, G is closed under inverses.

_{A}consisting of all the permutations f of A such that f(x) is contained in B for every x in B. Prove that G is a subgroup of S

_{A}.

- Let g be a permutation contained in G. If x is in B, then g(x) is in B. Since g(x) is in B, then f(g(x)) is in B as well. Similarly you can prove that [gof](x) is in B. Therefore, G is closed under composition.
- Since the images of all the elements of B are in B and permutations are bijective, then f
^{-1}carries elements of B to B, too. Therefore, G is closed under inverses.

**This is where I got stumped. The only ideas I have are these:**

B could be infinite which could affect the outcome. If B was infinite, then it might be possible for an element outside of B to have an image in B. Although that wouldn't make any sense since G is bijective. I cannot for the life of me think of a counterexample where either [fog] or f

^{-1}

*isn't*in B, especially since permutations are bijective. I'm a little muddy on how A being infinite would change anything.

**Knitting News?**Well, I'll save that for next time since I'm willing to bet that no one is reading this anyway ;-)

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