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+18 votes

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

  1. $p*q=q*p$ and
  2. $q*q=q$
in Set Theory & Algebra by Veteran (52.2k points)
edited by | 1.2k views

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3 Answers

+21 votes
Best answer
$p*p = q$
$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$

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).
by (327 points)
selected by
Thanks . explanation looks great.
Good observation!!!

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},∗)

Data given: p*p=q

CASE 1: Assume p∗q=p
Now, let's left multiply LHS and RHS with p, then we have
LHS = p*p*q
    = (p*p)*q      (using associative property)
    = (q)*q        (using given data: p*p=q)
RHS = p*p
    = q            (using given data: p*p=q)
Equating LHS and RHS, we have q*q=q

CASE 2: Assume p∗q=q
Now, let's left multiply LHS and RHS with p, then we have
LHS = p*p*q
    = (p*p)*q      (using associative property)
    = (q)*q        (using given data: p*p=q)
RHS = p*q
    = q            (using CASE 2 assumption)
Equating LHS and RHS, we have q*q=q


+9 votes
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.
by Veteran (60.9k points)
@Habibkhan ?
+3 votes
Cayley Table according to the question
  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
by Boss (33.9k points)
edited by
the question is very easy if we know that p*q=e but please explain how did u get this
This is semi group ,bt its not mentioned in the qs that is is a group .then how can we make cayley table?
how p*q =e ?
thnx got it

$e$ 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.

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

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