Prove the following function set is a group under function composition. Let $f_a:\mathbb{R} \to \mathbb{R}, f_a(x) = x\cosh(a) + \sqrt{1+x^2}\sinh(a)$ and the set $\mathbb{F}=\{f_a|a\in\mathbb{R}\}$. Prove $(\mathbb{F, \circ)}$ is a group.
I know $f_a\circ f_b = f_{a+b}(x)$, but I don't know how to prove it. I tried doing $f_a(f_b(x))$ but I didn't get anywhere.
 A: Set $$x:=\sinh u \tag{1}$$ 
($x \leftrightarrow u$ is a bijective correspondence from $\mathbb{R}$ to $\mathbb{R}$).
Then relationship 
$$f_a(x) = x\cosh(a) + \sqrt{1+x^2}\sinh(a)$$
becomes :
$$f_a(\sinh(u))=\sinh(u)\cosh(a)+\cosh(u)\sinh(a)$$
$$\iff \ f_a(\sinh(u))=\sinh(u+a) \tag{2}$$
As a consequence of relationship (2) (used twice):
$$\forall u \in \mathbb{R}, \ \ (f_a \circ f_b)(\sinh(u))=f_a(f_b(\sinh(u))=f_a(\sinh(u+b))=\sinh(u+a+b)=f_{a+b}(\sinh(u))$$
which is equivalent, due to bijective relationship (1), to 

$$\forall x \in \mathbb{R}, \ (f_a \circ f_b)(x)=f_{a+b}(x)\tag{3}$$

Relationship (3) confers a group structure to functions $f_a$ for composition by transfer of the additive group structure of $\mathbb{R}$.
No need to check group axioms one by one, but of course you can do it for fun using (3) everytime. For example, check that $f_0$ is the neutral element (you can even remark that it is a "natural neutral element"), that $f_{-a}$ is the reciprocal function of $f_a$...
Moreover, we have commutativity. Therefore, this group is abelian. This is exceptional for a group of functions !
Important remark : Relationship (2) can be expressed in a different way: 
$$f_a \ \circ \ \sinh \ = \ \sinh \ \circ \ t_a \tag{4}$$
where $t_a$ is the function defined by $t_a(x):=x+a$.
As a consequence, (4) can be written under the pretty "conjugated" form :

$$f_a \ = \ \sinh \ \circ \ t_a \ \circ \ \underbrace{\sinh^{-1}}_{\text{alias  argsinh}}\tag{5}$$

From (5), one deduces that for any $a$, $f_a$, being a composition of bijective functions, is itself bijective.
Remark: functions $f_a$ are solutions of the first order differential equation :

$$f'(x)=\sqrt{\dfrac{1+f(x)^2}{1+x^2}} \tag{6}$$ 

A particular case $f_0(x)=x$ gives $f_0'(x)=1$.
I don't resist a final pleasure : the display of the curves of some of these functions $f_a$, even if it is of no use for the answer to this question.

Fig. 1 : Curves of some functions $f_a$, for $a=-1$ (bottom curve) to $a=1$ (top curve) with step $0.2$. These curves are branches of hyperbolas (excepted the case of $f_0$ which is a straight line...)
