GATE CSE 1993 | Question: 28

Question

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

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

Answer 1

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).

Comments

Awesome proof!

---

@meghashyamc 

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?

Answer 2

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.

Comments

@Habibkhan ?

Answer 3

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

Comments

is this correct answer?

---

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

---

I think, this is not correct answer

Answer 4

A) Given that p*p=q

And wkt semigroup satisfy associativity

Therefore

p*(p*p)=(p*p)*p

P*q=q*p

Proved

 

B) plz explain to me