$A.$ TRUE
Define a set $S_{3}=$ Set of all bijections from $\{1,2,3\}$ to $\{1,2,3\}$
$f_{1}: 1\rightarrow 1, 2\rightarrow 2,3\rightarrow 3$
$f_{2}: 1\rightarrow 2, 2\rightarrow 1,3\rightarrow 3$
$f_{3}: 1\rightarrow 3, 2\rightarrow 2,3\rightarrow 1$
$f_{4}: 1\rightarrow 1, 2\rightarrow 3,3\rightarrow 2$
$f_{5}: 1\rightarrow 2, 2\rightarrow 3,3\rightarrow 1$
$f_{6}: 1\rightarrow 3, 2\rightarrow 1,3\rightarrow 2$
$S_{3}=\{f_{1},f_{2},f_{3},f_{4},f_{5},f_{6}\}$
Consider the composition of functions on $S_{3}$
Few Ex: $f_{2}(f_{5})=f_{4}$ $f_{5}(f_{2})=f_{3}$
Now,
1. $S_{3}$ is closed under composition
2. There is an identity element of $S_{3}:f_{1}$
3. Is there an inverse for every element of $S_{3}?$ Yes!!
$f_{2}(f_{2})=f_{1},f_{3}(f_{3})=f_{1},f_{4}(f_{4})=f_{1},f_{5}(f_{6})=f_{1}$
4. Composition of functions is associative.
$B.$ TRUE
Go through the Best answer.
$C.$ FALSE
Inverse is not possible $\times$
$D.$ TRUE
First assume that H is a subgroup of G. We wish to show that $gh^ {−1} ∈ H$ whenever g and h are in H. Since h is in H, its inverse $h^{−1}$ must also be in H. Because of the closure of the group operation, $gh^{−1} ∈ H.$