Is $G$ an Abelian group? Denote by $f_{a,b}$ the map from $\mathbb R$ to $\mathbb R$ such that $f(x)=ax+b$ for all $x \in \mathbb R$, where $a,b \in \mathbb R$ with $a\neq0$.
Let $G=\{f_{a,b}|a,b\in \mathbb R,a \neq0\}$. So I have already shown that $f_{a,b}$ is a bijection for any $a,b \in \mathbb R $ with $a\neq 0$ and prove that it is a subgroup of the group Sym$_{\mathbb R}$ for all bijective maps from $\mathbb R$ to $\mathbb R$. 

Now I want to find if $G$ is an abelian group or not. 
  I am aware abelian is the same as commutative, i.e., $ab=ba$. 

However I am confused as to how I would apply this to my group. This is a group under composition, or at least that's was I was required to use to show it is a subgroup. 
I assume the same still stands to show whether or not it is commutative. In any case I have tried two methods, both of which could be wrong.
I tried to show:
$$(ax+b)(a'x'+b')=(a'x'+b')(ax+b),$$
but under composition it would be trying to show:
$$(a'(ax+b)+b')=(a(a'x'+b')+b).$$ 
But they both don't seem right to me, in any case I'm not really sure what the right direction is. Any help/hints would be very appreciated. 
 A: A group $G$ is abelian iff $gh = hg$ for all $g,h \in G$. Now, what are elements in your group? They are functions. So let us take two of them $g = f_{a,b}$ and $h = f_{c,d}$. Now calculate $gh$ and $hg$:
$$ gh(x) = g(cx + d) = a(cx + d) + b = acx + ad + b $$
$$ hg(x) = h(ax + b) = c(ax + b) + d = cax + cb + d $$
Now two functions are equal iff their value is same on every $x$. Now can you say whether $gh = hg$?
A: Is $\;f_{a,b}\circ f_{c,d}=f_{c,d}\circ f_{a,b}\;$ ? Meaning:
$$a(cx+d)+b\stackrel?=c(ax+b)+d\iff acx+ad+b\stackrel?=acx +bc+d$$
Now you produce a counterexample to deduce $\;G\;$ is non-abelian.
A: It is about the second equation you posted (i.e. composition). But there is only one $x$ (no $x'$). Multuiplying it out then gives $$a'ax+a'b+b'=aa'x+ab'+b$$ which is not true in general. So no, the group is not abelian.
A: You have to take 2 elements in $G$, say, $g$=$f_{a,b} $ and $g'$=$f_{a',b'}$, now check the composition:
$g○g'(x)= f_{a,b} (a'x+b') = aa'x+ab'+b$
$g'○g (x)= f_{a',b'} (ax+b) =a'ax+a'b+b'$ 
From here is easy to construct a counterexample and show it is not abelian.
