Prove $G = ${bijections $f : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} $} is a group Let $G  = ${bijections $f : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} : f$ is continuous and $f^{-1}$ is continuous}.
Note: assume that the composition of continuous maps is continuous
Prove that $G$ is a group, and also provide an example of a subgroup $H$ of $G$ other than $H = ${$e$} or $H = G$.
I tried:
i) closure: I proved that a bijection composes with another bijection is a bijection.
ii) associativity: composition of functions is associative.
iii) inverse: I am stuck on this one. How to prove that an inverse of a bijection is also a bijection. And what is the identity element in this group?
iv) identity: what is the identity element in this group?
What is a good example of a subgroup of this group other than $H = ${$e$} or $H = G$?
Thank you.
 A: I will try to give a good framework of what you yourself should prove.
$1)$ Closure: Your notion here is correct, a composition of bijections is necessarily a bijection.
$2)$ Associativity: Once again, your notion is correct. Composition of continuous functions is an associative operation.
$3)$ Inverse: Can you prove that if $f:\mathbb{R}^2\to\mathbb{R}^2$ is a continuous bijection, then $f^{-1}$ is also a continuous bijection? It is actually pretty intuitive that the inverse mapping of a bijection is a bijection. Here is a small example. Consider the sets 
$$ X=\{1,2,3\},Y=\{-1,-2,-3\}.$$
Define the bijection 
$$\phi:\{1,2,3\}\to\{-1,-2,-3\}: \phi(x)=-x, \forall x\in X.$$
The inverse function is clearly
$$ \phi^{-1}:\{-1,-2,-3\}\to\{1,2,3\}:\phi^{-1}(y)=-y,\forall y\in Y.$$
$\phi^{-1}$ is obviously a bijection. Can you generalize this?
$4)$ The identity function on a set maps every element to itself. Is this not a bijection?
An easy construction of a subgroup is a group of mappings that fixes all but finitely many points, and permutes the others. I'll leave this as an exercise.
A: The identity element is the identity map, and this also serves as an example of a subgroup (the trivial subgroup). To show that the inverse is also a bijection, use the fact that every element in $\mathbb {R}^2$ has exactly one unique image under any $f$ in your group.
A: The identity element of the group is just the identity map. I'm confident you can prove that the inverse of a bijection is also a bijection yourself (perhaps you could start by computing some examples of inverses of bijections from finite sets to themselves, and show that the inverse of those bijections are both surjective and injective).
For a subgroup, you're essentially looking for a property $P$ of continuous bijections such that the identity has $P$, if $f,g$ both have $P$, then $f\circ g$ has $P$, and if $f$ has $P$, then $f^{-1}$ has $P$.
Some candidates for $P$: has fixed points, doesn't have fixed points, is differentiable, is an isometry, is an additive homomorphism, is linear, is a rotation, is linear and diagonalizable, is not linear, is not a rotation,...etc
A: What you are asking for is essentually that $\text{Aut}(\mathbb{R}^2)$ is a group.


*

*Yes, composition of isomorphisms is an isomorphism which is fairly easy to show

*Not hard either if just cumbersome

*To prove that the inverse of an isomorphism $\varphi$ is an isomorphism we just need to show that it is an epimorphism and monomorphism, for monomorpshim we just show that $\ker\varphi = 0 = (0,0)$. We have $\varphi(0)=0$ and for any $a\neq 0$ that $\varphi(a)\neq 0$, by that $\varphi$ is an isomorphism. This means that $\varphi^{-1}(0)=0$ because our $\varphi$ only took $0$ to $0$. So it is a monomorphism, next we let $x,y\in\mathbb{R}^2$ be given and have that $y=\varphi(x)$, this means that $\varphi^{-1}(y)=x$ so for any $x$ there exists a $y$ thereby making it an epimorphism and thereby an isomorphism and in turn an automorphism.

*$\text{id} x = x$ is the identity element.

