The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+1 vote
149 views

How to solve this type of questions??

 

asked in Set Theory & Algebra by Boss (36.4k points)
edited by | 149 views
0

@Lakshman Patel RJIT,

Please add test-series name in title

0
See ,now i add the test series name
+1
simple approach to check whether a group is abelian or not is

if the order of group is prime square then the group is abelian

here order of group 4 (that is 2 square)

1 Answer

+4 votes
Best answer
First, you need to check if given group is abelian or not. Here associativity, the existence of an identity, existence of inverse for each element and commutativity are all satisfied so the group is abelian. Now, the order of an element is a smallest positive integer n such that a^n=e. for element a, a^2= a*a=b,  a^3= a^2 * a= b*a=c,  a^4= a^3*a=c*a=e.

Hence the order of a=4. Option C is correct!
answered by Loyal (7.6k points)
selected by
+1
well explained !!
0
thanks!! :)
0

a, a^2= a*a=b,  a^3= a^2 * a= b*a=c,  a^4= a^3*a=c*a=e.@Utkarsh Joshi I didn't understand this line

 

+1

@

If $'a'$ is the order of the group, then for smallest $'n'$ we get $a^{n}=e$  where $'e'$ is the identity element.

$(a)^{1}=a$

$(a)^{2}=a*a=b$

$(a)^{3}=a*a*a$

$'*'$ is associative.

So, we can write like this

 $(a)^{3}=(a*a)*a=b*a=c$

          $(or)$

 $(a)^{3}=a*(a*a)=a*b=c$

$(a)^{4}=a*a*a*a=(a*a)*(a*a)=b*b=e$

                   $(or)$

$(a)^{4}=a*a*a*a=a*(a*a*a)=a*c=e$

                    $(or)$

$(a)^{4}=a*a*a*a=(a*a*a)*a=c*a=e$

So,order of group$=4$

$\Rightarrow$Order of group$=$Order of generator$=4$

0
@utkarsh @lakshman

How it's inverse and associative ?

How order of element is a^n?

Please explain

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,434 questions
53,630 answers
186,007 comments
70,899 users