Let ,  ∈ ℝ and let $_{,}$: ℝ → ℝ be the function defined as $_{,}$ () =  +  for all  ∈ ℝ. Use this for the problems below. a) Prove that  = {$_{,}$ | ,  ∈ ℝ,   ≠ 0} is a group, where the operation is composition.
Let 1 = {$\big(\begin{smallmatrix}
  a & 0\\
  b & 1
\end{smallmatrix}\big)$  ∶ ,  ∈ ℝ,  ≠ 0}, where the operation is matrix multiplication, and let 
be the group in part a).\
b.)Prove that the map :  → 1 defined as (,) = $\big(\begin{smallmatrix}
  a & 0\\
  b & 1
\end{smallmatrix}\big)$  for all $_{,}$ ∈ , is an isomorphism.
Proof:
a) the associative property is always true for composition of functions.
I am not sure how to do the other two properties for group to show that there is an inverse and identity exist. This is preventing me to do part b if someone can help guide me through this proof. I need help formulating the rest of the proof.
 A: Let $a,b,c,d\in\mathbb{R}$ be arbitrary elements, $a,d\neq 0$. Observe that for all $x\in \mathbb{R}$, $$f_{a,b}\circ f_{c,d}(x)=f_{a,b}(f_{c,d}(x))=f_{a,b}(cx+d)=a(cx+d)+b=(ac)x+ad+b=f_{ac,~ad+b}(x),$$ i.e., $f_{a,b}\circ f_{c,d}=f_{ac,~ad+b}$, this means that the composition is closed on $G$. Now, $f_{1,0}$ is the identity of $\circ$. In fact, $$f_{a,b}\circ f_{1,0}(x)=f_{a,b}(f_{1,0}(x))=f_{a,b}(x)=ax+b=f_{1,0}(ax+b)=f_{1,0}(f_{a,b}(x))=f_{1,0}\circ f_{a,b}(x),$$ for all $x\in \mathbb{R}$. Now we will calculate $f_{a,b}^{-1}$. We want to found a map $f_{r,s}\in G$ such that $$f_{a,b}\circ f_{r,s}=f_{r,s}\circ f_{a,b}=f_{1,0}.$$ For all $x\in \mathbb{R}$, $f_{a,b}\circ f_{r,s}(x)=f_{a,b}(rx+s)=a(rx+s)+b=arx+s+b=f_{1,0}(x)=x$ if and only if $ar=1$ and $s+b=0$, that is, $r=1/a$ and $s=-b$. It means that $f_{1/a,~-b}$ is a right inverse to $f_{a,b}$ on $G$. You can check that $f_{1/a,~-b}$ is a left inverse to $f_{a,b}$ on $G$ and so it is an inverse to $f_{a,b}$. In particular, every element of $G$ has an inverse because $a,b\in \mathbb{R}$, $a\neq 0$, are arbitrary. Finally, as you said, the composition is always associative. Therefore, $G$ is a group.
Your map $\phi$ cannot by an isomorphism because it is not even a homomorphism: it is easy to check that $\phi(f_{1,2}\circ f_{2,1})\neq \phi(f_{1,2})\phi(f_{2,1})$. Now, if you define $$G_{1}=\{A=[a_{ij}]\in M_{2\times 2}(\mathbb{R});~a_{11}\neq 0,~a_{21}=0~\text{and}~a_{22}=1\}$$ we will show that $G$ and $G_{1}$ are isomorphics. Consider $\phi:G\longrightarrow G_1$ defined by $\phi(f_{a,b})=\left[ {\begin{array}{cc}
   a & b \\
   0 & 1 \\
  \end{array} } \right].$
Observe that $$\phi(f_{a,b}\circ f_{c,d})=\phi(f_{ac,ad+b})=\left[ {\begin{array}{cc}
   ac & ad+b \\
   0 & 1 \\
  \end{array} } \right]=\left[ {\begin{array}{cc}
   a & b \\
   0 & 1 \\
  \end{array} } \right]\left[ {\begin{array}{cc}
   c & d \\
   0 & 1 \\
  \end{array} } \right]=\phi(f_{a,b})\phi(f_{c,d}),$$ and so $\phi$ is a homomorphism. Clearly $\phi$ is a surjective map and we just have to show $\phi$ is injective. But, if $f_{a,b}=1_{G_{1}}$, then $a=1$ and $b=0$, i.e., $f_{a,b}=f_{1,0}$. It means that $\phi$ is injective and so it is a isomorphism between $G$ and $G_1$.
A: We need to show that $G$ is closed under composition. Let $f_{a,b}, f_{c,d} \in G$. Show that $f_{a,b} \circ f_{c,d} \in G$.
Hint: We see $f_{a,b} \circ f_{c,d}(x) = f_{a,b}(cx + d) = a(cx+d)+b = \cdots.$
We need to show that there is an identity. The identity for composition of functions is a function such that $f_{a,b}(x) = x$ for all $x$ (why?). Determined $a$ and $b$ which will give us this.
Hint: We want $f_{a,b}(x) = ax+b = x,$ so$\ldots$
Hint: To see this is an identity, notice $f_{a,b} \circ f_{c,d}(x) = \cdots$ for the above $a$ and $b$.
We need to show existence of inverse. Fix $f_{a,b} \in G$. In the first part, we calculated $f_{a,b} \circ f_{c,d}$. What conditions do we need on $c$ and $d$ so that $f_{a,b} \circ f_{c,d}(x) = x$ for all $x$? For the same $c$ and $d$, do we have $f_{c,d} \circ f_{a,b}(x) = x$?
This shows that $G$ is a group. (Assuming you know associativity)
