.#1 a) If ex = x for some elements e,x belong to S, we say e is a left identity for x; similarly, if xe = x we say e is a right identity for x. Prove that an element is a left identity for one element of S if and only if it is a left identity for every element of S Let S be a non-empty set with a binary operation which is associative and both left and right transitive
b) Prove that S has a unique identity element
c) Deduce that S is a group under the given binary operation
#2.Prove that n|φ(a^n-1) for every integer a≥2 and any positive integer n
Transitivity is the key - essentially, this means that for all there exists a such that and (or certainly that is my understanding of it applied here - essentially you are taking the action of the semigroup over the semigroup and saying it is transitive).
So, if for some element we have then that and so it is a left identity for each element in .
The proof is analogous for the right identity, and then the result follows from Defunkt's post.
@Defunkt - Hi-Thanks for your post. I am in doubt if I am following what you wrote. So, let me plz state that and you tell me if I got it correct or not
S - is a set with a binary operation defined
e1 - Right Identity of the set => x.e1 = x, for all x in S
e2 - Left Identity of the set => e2.x = x, for all x in S
Are we trying to show e1=e2=e (unique identity element in S)?