ISRO2018-17

Question

$(\text{G}, \ast )$ is an abelian group. Then

1. $x= x^{-1}$ for any $x$ belonging to $\text{G}$
2. $x=x^{2}$ for any $x$ belonging to $\text{G}$
3. $\left( x \ast y \right )^{2}= x^{2} \ast y^{2}$ , for any $x, y$ belonging to $\text{G}$
4. $\text{G}$ is of finite order

Comments

No, it's correct.

Share a resource if you don't think it's correct.

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@'JEET

Please edit your above table

@Hradesh patel Sir You are right.

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Done thanks.

Answer 1

A) FALSE; if every element in the group is its own inverse, i.e $x = x^{-1}$ for every $x\ \epsilon\ G\ \implies\ G\ is\ Abelian$ but the converse is not true. 

$( \mathbb{Z},\ +) $ is an infinite abelian group but the following are not true for it:

e = 0

$ x = x^{-1}$: Inverse of 1 $\neq$ 1 but  = -1

B) FALSE; 

$1^2 = 1 + 1 \neq 1\ but = 2$

C) TRUE; G is abelian if and only if $(xy)^2 = x^2y^2$ for all x and y in G.

($\Rightarrow$)

Let x, y ∈ G. Then $(xy)^2 = xyxy = xxyx$ since G is abelian (COMMUTATIVE), and of course $xxyy = x^2y^2$. Thus $(xy)^2 = x^2y^2$.

($\Leftarrow$)

Assume $(xy)^2 = x^2y^2$ for all x, y ∈ G. Then we have the following:

$(xy)^2 = x^2y^2 \\xyxy = xxyy \\x^{−1}xyxy = x^{−1}xxyy \\yxy = xyy \\yxyy^{−1} = xyyy^{−1} \\yx = xy$

Thus G is abelian.

D) FALSE; Since $( \mathbb{Z},\ +) $ is an infinite abelian group.

Comments

Can you please explain how option 'C' is true for (Z, +).

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Please consider the below scenario and explain it:

(G,*) is (Z,+)

then (2+3)^2 will not  be equal to 2^2+3^3

so, how option C can be true?

we can not ignore the possibility of  (G,*) being as (Z,+).

Please, anyone, explain this solution.

Thanks in advance.

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You are doing it wrong!

See here

In ($\mathbb{Z},+), (xy)^2=x^2y^2$ means $2(x+y)=2x + 2y$.

Now coming to your question,

$(2+3)^2=(2+3)+(2+3)=(2+2)+(3+3)=2^2+3^2$

So, yes it holds true!

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Option (B) doesn't say that $x^2=x*x$. Why did you do $1^2=1+1$?

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Option B: x=x^2 for any x belonging to G

so, if we take x=1

1=1^2

i.e, 1=1+1   (as (G,+) taking * as  + operator as * can  be taken as any operator)

1=2 that is false thus option B is not true

Answer 2

option C is correct.

take any two elements a,b from the group

for C,

for abelian group the following property must hold:

    b*a=a*b

=>a*(b*a)*b = a*a*b*b

=>(a * b)*(a * b)=(a*a)*(b*b)(commutative property of group)

=>(a * b)² = a² * b²

Comments

Sambit for the proof of option A you are showing that if x=x^(-1) then (G,*) is abelian.

You have to prove if (G,*) is abelian then x=x^(-1)

And you are doing the same thing for option C.

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oh sorry.i actually didn't notice that.

Answer 3

The statement of option C is (x*y)^2=x^2*y^2. when (G,*) is an abelian group.

Now, (x*y)^2 = (x*y)*(x*y)= x*y*x*y

Since (G,*) is abelian we have

x*y*x*y = x*x*y*y = x^2*y^2 (proved)

So, option C is correct.

Rest of the options are incorrect.

Comments

Can you please clarify whether '*' could be any operator or is it the Multiplication operator only?

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* is any operator

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So if we consider the case of '*' being addition operator(as it will also form an Abelion Group) then how would (a*b)²=a²*b² ?