Show semigroup ie associativity Show monoid ie identity This

Show semigroup (i.e. associativity)

Show monoid (i.e. identity)

This is an abstract algebra proof. Please explain and show all working. Thank you so very much!!!

Solution

Semi Group

Let X be a non-empty set and B(X) denote the set of all binary relations on X.
Define a binary operation ‘’ on B(X) as follows:
If , B(X), then = {(x,y) X × X : if z X such that (x,z) and (z,y) }.
Then B(X) is a semigroup with identity element = {(x,x) : x X}.

A relation on a semigroup S is called a congruence relation on S iff

(i) is an equivalence relation on S;
(ii) (a,b) (ca,cb) and (ac,bc) for all c S.

Let be a congruence relation on a semigroup S and S/ the set of all -equivalence classes a,a S.

Define a binary operation ‘’ on S/ by

a b = (ab), for all a,b S. Clearly, the operation ‘’ is well defined.
Then (S/,) becomes a semigroup.

Let G be a semigroup with an identity element 1

Let b and b be inverses of a in G, then

b = b1
= b(ab), since b is an inverse of a.
= (ba)b (by associativity in G)
= 1b, since ba = 1
= b.

We denote by a1 the unique inverse (if it exists) of an element a of a semigroup.
The set M(X) of all mappings of a given non-empty set X into itself, where
the binary operation is the usual composition ‘’ of mappings, forms an important
semigroup.

Monoid

Clearly, (M(X),) is a semigroup under usual composition of mappings. Let
1X : X X be the identity map defined by

1X(x) = x for all x X.

Then for any f M(X),

(1X f )(x) = 1X(f (x))

                = f (x) and (f 1X)(x) = f (1X(x)) = f (x)
for all x X imply 1X f = f 1X = f . Consequently,

M(X) is a monoid with 1X as its identity element.

Every element in M(X) may not have an inverse.
For example, if X = {1, 2, 3}, then f M(X) defined by

f (1) = f (2) = f (3) = 1 has no inverse in M(X). Otherwise, there exists an element g M(X)
such that f g = g f = 1X. Then

1 = 1X(1) = (g f )(1) = g(f (1)) = g(1) and
2 = 1X(2) = (g f )(2) = g(f (2)) = g(1) imply that g(1) has two distinct values.
This contradicts the assumption that g is a map.
We now characterize the elements of M(X) which have inverses.