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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. $q*q=q$
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This might help ...

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, $q*q = LHS = p*p*p*p = p*p = q = RHS. ($From $(1)).$

CASE 2: $q = p*q.$
This means, $q=p*q = p*p*p\quad \to (2)$
Then, $q*q = LHS = p*p*p*p = p*q = q = RHS$ (based on Case 2's assumption).
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Thanks . explanation looks great.
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Good observation!!!
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plz explain this line, i am not getting this

This means, either p=p∗qp=p∗q or q=p∗qq=p∗q as the semi-group is ({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.
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@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
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the question is very easy if we know that p*q=e but please explain how did u get this
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This is semi group ,bt its not mentioned in the qs that is is a group .then how can we make cayley table?
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how p*q =e ?
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thnx got it
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$e$ is not even part of semigroup, semigroup has only two elements $p$ and $q$.
How are u doing this?

moreover, Cayley Table is defined for Groups only, not for semigroup.

Correct me if wrong.

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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
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this is a semigroup. how can we make cayley table ?

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