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.