${\color{Red} All}$ are ${\color{Magenta}True }$
a) In a group (G,*), if a*a = a, then a=e, where e is an identity element.
We will prove this by contradiction, $(a*a = a) \rightarrow a = e$
let us assume that $a \neq e$ is true
that means a is not the identity element
According to our assumption there must exist some identity element (e) that satisfies
a*e1 = a or e2 * a = a
e1 = e2 = e ( we know that * is commutative and associative)
and we assumed that $a \neq e$ but if $(a*a = a)$ is true then $a \neq e$ can't be true because
$(a*a = a)$ a acts as both e1 and e2
b) In a group (G,*), if x-1 = x, ∀x ∈ G, then G is an Abelian group.
An abelian group is a group in which the law of composition is commutative.
option b says that every element if set G is inverse of itself. This is TRUE
Assume G = {−1,1}
c) In a group (G,*), if (a*b)2 = (a2*b2) then G is an Abelian group.
assume G be any set of integers
Let G = {1,2},
(1*2)2 = (12 * 22)
d) In a group (G,*), if (a*b)n = (an*bn) then G is an Abelian group, where n={2, 3, 4, ......}
Same reasoning as option c