a. Lets assume a*a!=a

So, if our assumption is true then, (a*a)*a !=a*(a*a) [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

but , (a*a)*a=a*(a*a) because (A,∗) is a semigroup hence associative, which proves that our initial assumption is wrong,

hence a*a=a

b.Lets assume a∗b∗a!=a

So, if our assumption is true then , (a*b*a)*a!=a*(a*b*a) [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

Now,(a*b*a)*a= a*b*(a*a) [since (A,*) is associative]

or, (a*b*a)*a=a*b*a [since a*a=a (already proved) ]------------------ (i)

Now, a*(a*b*a)=(a*a)*b*a [since (A,*) is associative]

or, a*(a*b*a)=a*b*a [since a*a=a (already proved) ]-------------------(ii)

using (i) and (ii) (a*b*a)*a= a*(a*b*a) , hence proved that our initial assumption was wrong hence,a∗b∗a=a

c. Lets assume a∗b∗c !=a∗c

So, if our assumption is true then , (a*b*c)*(a*c)!=(a*c)*(a*b*c) [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

Now,(a*b*c)*(a*c)= a*b*(c*a*c) [since (A,*) is associative]

or, (a*b*c)*(a*c)= a*b*c [since (c*a*c)=c ( for any a,b , a*b*a=a already proved as part of question b) ]------------------ (i)

Now, (a*c)*(a*b*c)=(a*c*a)*b*c [since (A,*) is associative]

or, (a*c)*(a*b*c)=a*b*c [since (a*c*a)=c ( for any a,b , a*b*a=a already proved as part of question b) ]-------------------(ii)

using (i) and (ii) (a*b*c)*(a*c)=(a*c)*(a*b*c) , hence proved that our initial assumption was wrong hence,a∗b∗c =a∗c