plz with details 4. Recall \( S(\mathbb{R}) \) is the group of all permutations of \( \mathbb{R} \) under the composition of functions. For any pair of real numbers \( a \neq 0 \) and \( b \), define a function \( f_{a,b} : \mathbb{R} \to \mathbb{R} \) as follows:\[ f_{a,b}(x) = ax + b \]a. Prove that \( f_{a,b} \in S(\mathbb{R}) \). That is, \( f_{a,b} \) is a permutation of \( \mathbb{R} \).b. \( f_{a,b} \circ f_{c,d} = f_{ac,ad+b} \), where \( f_{a,b} \circ f_{c,d} \) is the composition of \( f_{a,b} \) and \( f_{c,d} \).c. \( (f_{a,b})^{-1} = f_{\frac{1}{a},-\frac{b}{a}} \), where \( (f_{a,b})^{-1} \) is the inverse of \( f_{a,b} \).d. Let \( H = \{ f_{a,b} \mid a \in \mathbb{R}, b \in \mathbb{R}, a \neq 0 \} \). Prove that \( H \) is a subgroup of \( S(\mathbb{R}) \).

given a group of all permutation of ℝ that is Sℝ , for any pair of real number a≠0 , b define a function fa,b:ℝ→ℝ by fa,bx=ax+b for a≠0 , b(a) to show : fa,b is a permutation of ℝ we will show it is one-one and onto (1) one-one for fa,bx1=fa,bx2⇒ax1+b=ax2+b⇒ax1=ax2⇒ax1-x2=0⇒x1=x2 ∵a≠0 that is fa,b is one-one(2) onto for any y∈ℝ , we will find x such that fa,bx=y⇒ax+b=y⇒ax=y-b⇒x=1ay-b a≠0 ∴1aexist thus , we have that fa,b is a permutation in ℝ (b) to show : fa,b∘fc,d=fac,ad+b where fa,b∘fc,d is composition of fa,b and fc,d fa,b∘fc,dx=fa,bfc,dx =fa,bcx+d ∵fa,bx=ax+b =acx+d+b =acx+ad+b =fac,ad+b hence proved