Prove that the set of one-to-one function $f: S\to S$ is a group with composition as group operation. Let $S$ be any set.
a) Prove that the set of one-to-one function $f: S\to S$ is a group with composition as group operation.
b) let $f$ and $g$ are any one-to-one function from $S$ onto $S$. Is it necessarily true that $f\circ g=g\circ f$
for a) I call $W$ is the set of functions $f$. I know that I need to show that $W$ is closed and associative under its operation, $W$ contain an identity element, $W$ contain an inverse for each of its element. But I don't know how to explain how I got these conditions.
For b) I don't think it's true, but I haven't found any counter example yet.
 A: A function $f$ is injective iff $f(a)=f(b)$ implies $a=b$, for every pair of elements $a$ and $b$. Now suppose $(f\circ g)(a)=(f\circ g)(b)$. By definition of $\circ$, you have $(f\circ g)(a)=(f\circ g)(b)$ implies $f(g(a))=f(g(b))$ which implies (use injectivity of $f$) that $g(a)=g(b)$ which implies (now use injectivity of $g$) that $a=b$. Since composition is associative for an arbitrary function, it is associative also for the subset of functions given by injective ones. For the identity, note that the identity function is trivially injective.
For the inverse, you need also surjectivity (an injection of a set in itself is not in general also surjective). You want to show that $f^{-1}$ is also one-to -one. So suppose $f^{-1}(a)=f^{-1}(b)$. Apply $f$ to both sides and deduce $a=b$. Here you use injectivity to guarantee that $f^{-1}$ is defined.
A: For a) (assuming as the comments mention you also add surjectivity), perhaps you could write out explicitly which parts you are confused about. It should be an exercise in writing out and checking definitions.
For b), let $S = \mathbb{R}$ and put $f(x) = 2x$, $g(x) = x+1$.
