Define the functions $f_{a,b}(x)=ax+b$, where $a, b \in \R$ and $a>0$.

Show that $G:=\{ f_{a,b} \mid a, b \in \R, a>0\}$ is a group . The group operation is function composition.

Steps.

Check one by one the followings.

1. The group operation on $G$ is associative.
2. Determine/guess the identity element and show that it is in fact the identity element.
3. Determine/guess the inverse of each element and show that it is in fact the inverse.

Proof.

The product of $f_{a_1, b_1}$ and $f_{a_2, b_2}$ is given by
\[ f_{a_2, b_2}(x)\circ f_{a_1, b_1}(x) =a_2(a_1x + b_1)+b_2=a_2a_1 x +a_2b_1 + b_2.\] Since the group operation is function composition, it is associative.

The identity element is $f_{1,0}=x$ since for any $f_{a,b} \in G$, we have
\[f_{a,b}(x)\circ f_{1,0}(x)=af_{1,0}(x)+b=ax+b=f_{a,b}(x)\] and
\[f_{1,0}(x) \circ f_{a,b}(x)=x\circ f_{a,b}(x)=f_{a,b}(x).\]

Now we find the inverse of $f_{a_1,b_1}$.
If $f_{a_2, b_2}(x)\circ f_{a_1, b_1}(x)=x(=f_{1,0})$, then we should have $a_2 a_1=1$ and $a_2 b_1 +b_2=0$. Solving these, we obtain $a_2=1/a_{1}>0$ and $b_2=-a_2 b_1=-b_1/a_{1}$.
Thus our candidate of the inverse $f_{a,b}^{-1}$ is
\[f_{1/a_1, -b_1/a_1}=x/a_1-b_1/a_{1}.\]

In fact it is the inverse since it also satisfies
\[f_{a_1,b_1} \circ f_{1/a_1, -b_1/a_1}=a_1 (x/a_1-b_1/a_{1})+b_1 =x-b_1+b_1=x.\] Thus $f_{a,b}^{-1}=f_{1/a_1, -b_1/a_1}$ and we conclude that $G$ is a group.

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