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How to solve this type of questions??

 

in Set Theory & Algebra
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@Lakshman Patel RJIT,

Please add test-series name in title

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See ,now i add the test series name
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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)

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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!

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well explained !!
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thanks!! :)
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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

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