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## Homework Statement

Let T be a subset of S and consider the subset U(T)={f [itex]\in[/itex] A(S) | f(t)[itex]\in[/itex]T for every t[itex]\in[/itex]T}.

1) If S has n elements and T has m elements, how many elements are there in U(T)?

2) Show that there is a mapping F:U(T) -> S

_{m}such that F(fg)=F(f)F(g) for f,g[itex]\in[/itex]U(T) and F is onto S

_{m}

3) If m<n, can F ever be 1-1? If so, when?

## The Attempt at a Solution

1) m!*n! ?

2) I've already shown in the previous problem (Problem 18 in baby Herstein) that if f and g are in U(T), so is their product fg. Now, If I define F as a function that takes an element in U(T) and restricts the domain to only the elements in T and I denote it by F=f[T], then It's obvious that the mapping F=f[t] under composition satisfies F(fg)=F(f)F(g). because F(fg)=(fg)[T]=f[T]og[T]=F(f)F(g). To show that this mapping is surjective, for any element in S

_{m}like h one could define the element f={ f(n)=h(n) for every t in T & f(n)=n otherwise}, It's obvious that F(f)=h, hence, F is onto S

_{m}.

3) No, F is not injective. because If we assume T={1,2,3} and S={1,2,3,4,5,6} and we consider the mapping F:U(T)->S

_{6}and then we define f(1)=2 f(2)=3 f(3)=1 f(4)=4 f(5)=5 and g(1)=2 g(2)=3 g(3)=1 g(4)=5 g(5)=4, then both f and g are mapped to the same element in S

_{6}. However, we can define another map that is injective. we define f~g iff F(f)=F(g). we define the equivalence class of f as [f]={g| f~g} and we define the quotient set U(T)/~ as U(T)/~={[f]|f[itex]\in[/itex]U(T)}. We define [f][g]=[fg] and It's easy to show that this multiplication law is well-defined on U(T)/~ and it turns U(T)/~ into a group. then the mapping F': U(T)/~ -> S

_{m}is injective.

I guess We can also reform F in another way. We can form a subgroup of U(T) by choosing only those permutations that permute the elements in T to new elements in T and keep the rest as they are. This is a subgroup of U(T) that I like to denote it by K and It's obvious that composition of permutations is a well-defined operation over it. The mapping F'': K->S

_{m}is injective.

Well, I just wanna know whether my proof is correct or not.

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