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Let $\left(\{ p,q \},*\right)$ be a semigroup where $p*p=q$. Show that:

1. $p*q=q*p$ and
2. $q*q=q$

a.
$p*p = q$
$p*p*p = p*q \quad \quad$//left operation with $p$
$(p*p)*p = p*q \quad \quad$//associative property
$q*p = p*q \quad \quad\quad$//$p*p=q$

b.
For a semi-group, two properties are known: associativity and closure. (Identity is not required).
Closure means that $p*q$ must be a part of the semi-group.
This means, either $p=p*q$ or $q=p*q$ as the semi-group is $\left(\{p,q\},*\right)$

CASE 1: $p = p*q.$
This means, $p=p*p*p$ as $p*p = q \quad \to (1)$
Then, $q*q = \text{LHS} = p*p*p*p = p*p = q = \text{RHS}. ($From $(1)).$

CASE 2: $q = p*q.$
This means, $q=p*q = p*p*p\quad \to (2)$
Then, $q*q = \text{LHS} = p*p*p*p = p*q = q = \text{RHS}$ (based on Case $2's$ assumption).

Awesome proof!

case 2:

q = p * p *p

q * q = p* p* p *q ( I think you have done mistake here).

q*q = p *q

q*q = q

@meghashyamc

CASE 2: q=p∗q.

and in question  p*p=q

so using both we can write  p*p=p*q

and by left cancellation  p=q.

is it correct?

can q be both p*q and p*p at the same time? and p=q?

p*p = q
p*p*p = p*q         //left operations with p
(p*p)*p = p*q      //associative property
q*p = p*q            //p*p=q

m not getting 2nd part.

### 1 comment

@Habibkhan ?
 e p q e e p q p p q q q

[a] p*q = p*(p*p) = p*p*p = (p*p)*p = q*p (which should be e as each element should have an inverse)

[b] q*q = p*(p*q) = p*(q*p) = (p*q)*p = e*p = p.

So, the Cayley Table becomes:

 e p q e e p q p p q e q q e p

Cayley table describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplication table
this is a semigroup. how can we make cayley table ?