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$(\text{G}, \ast )$ 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
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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.

5 votes
5 votes

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

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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.
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