Proving $K$ is a group Now I have proved certain things are a group before, and I know that it requires:
1)Associativity
2)Inverse
3)Identity
But here I have such a strange thing that I wanted to clarify that I am doing it right.

If $a,b \in \mathbb{R}$ with $a\ne0$, let $L_{a,b}:\mathbb{R} \to \mathbb{R}$ be the function given by $L_{a,b}(x) = ax+b$ for all $x$. Let
$$K = \{L_{a,b}|a,b\in \mathbb{R}\}$$
Prove that $K$ forms a group under composition of functions.

So normally you want to test some operation binary operation like $x+y = x+y+2xy$ or something, and see if it meets the above $3$ requirements. But here I don't know if I want to look at associativity in which of the following ways:
1) $ ((ax+b) + (cy+d))+(ez+f) = (ax+b) + ((cy+d)+(ez+f))$
2) $ ((ax+b) + (cx+d))+(ex+f) = (ax+b) + ((cx+d)+(ex+f))$
3) $ ((ax+b) + (ay+b))+(az+b) = (ax+b) + ((ax+b)+(ax+b))$
4) $ ((ax+b) + (ax+b))+(ax+b) = (ax+b) + ((ax+b)+(ax+b))$
Which one am I meant to be comparing?
How do I get the Inverse and Identity our? Does $L_{a,b}$ have some $x$ such that $ax+b=ax+b$, so just $x=x$ for identity, and some $ax+b=x$?
 A: None of them. Rememeber the group operation is composition of functions, not pointwise addition.
You need to check $(L_{a,b}\circ L_{c,d})\circ L_{e,f}=L_{a,b}\circ (L_{c,d}\circ L_{e,f})$ as functions of $x$.
A: We need to remember we the group operation is function composition, and not function addition.  
So to establish associativity, we need to establish that $$(L_{a,b}\circ L_{c,d})\circ L_{e,f}=L_{a,b}\circ (L_{c,d}\circ L_{e,f})$$

Associativity
$$\begin{align} (L_{a,b}\circ L_{c,d})\circ L_{e,f} 
& = (L_{a, b}(cx + d))\circ L_{e, f}\\ 
& = [a(cx + d) + b] \circ L_{e, f}\\
& =[acx + ad+b]\circ L_{e, f}\\ 
& = L_{ac, ad+b}(ex + f) \\
&=ac(ex+f) + (ad+b) \\ 
&= a(cex+cf +d) + b\\
& =L_{a, b}\circ (cex + cf + d)\\  
& = L_{a, b} \circ [c(ex+f) + d ]\\  
&=L_{a, b} \circ(L_{c, d}( ex + f))\\ 
& = L_{a, b}\circ(L_{c, d} \circ L_{e, f})
\end{align}$$
Identity
$$L_{\text{id}}(x) = x = 1\cdot x + 0 = L_{1, 0}$$
Inverses
Let $L_{a, b}(x) = y =ax + b$ be any element in the group. Solve for $y$: $$y-b = ax \iff \frac{y-b}{a} = x$$
So $$L^{-1}_{a, b}(x) = \frac{x-b}{a} = \frac 1a\cdot x + \left(\frac{-b}a\right) = L_{\frac 1a, \frac{-b}a}$$
You can confirm that $L_{\frac 1a, \frac{-b}{a}}(x) = L^{-1}_{a, b}(x)$ by composing the two functions and obtaining the identity function: $L_{1, 0}$.
A: Your question essentially boils down to: "What do elements of $K$ look like?" The answer is obvious from the definition of $K$ and $L_{a,b}$. Each element in $K$ is a degree one polynomial with arbitrary coefficients and constants in $\mathbb R$. So, we see that 2) is the claim you wish to prove. Also, note that the operation defined on $K$ is function composition and not function addition.
