<s>[INST] For each of the following rules, either prove that it is true in every group $G$, or give a counterexample to show that it is false in some groups. Let $J$ be a group that consists of six matrices:
$I = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}, A = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 & 1 \\-1 & -1 \end{pmatrix}, C = \begin{pmatrix} -1 & -1 \\0 & 1 \end{pmatrix}, D = \begin{pmatrix} -1 & -1 \\1 & 0 \end{pmatrix}, K = \begin{pmatrix} 1 & 0 \\-1 & -1 \end{pmatrix}$.

$ \text { If $x^2 = e$, then $x = e$.}$

$A^2 = I.$ So the rule above doesn't hold in group $J.$

$\text { If $x^2 = a^2$, then $x = a$.}$

$A^2 = I = C^2$, but $A \neq C.$ So the rule above doesn't hold in group $J.$

$ (ab)^2 = a^2b^2.$

$BA = K, K^2 = I.$
$A^2 = I, B^2 = D, DI = D.$
So the rule above doesn't hold in group $J.$

$\text { If $x^2 = x$, then $x = e$}$.

Let $x^2 = x$. Then, $xxx^{-1} = xx^{-1}.$ So, $x = e.$ The rule above holds for all groups $G$.

$ \text { For every $x$ in  $G$, there is some $y$ in $G$ such that $x = y^2$}$.

Let $x = K$. Then, $K = y^2 = (\sqrt{K})^2 = K.$ So the rule above holds.

$ \text { For any two elements $x$ and $y$ in $G$, there is an element $z$ in $G$ such that $y = xz$}$.

Given: $xzx^{-1} = z.$ Let $x = A.$ Then $AzA \neq z.$ At least I wasnt able to show it. So, this rule fails.

Please, check my work.
 [/INST] The second to last statement is false. Consider for example the integers with multiplication.
The last statement is true. Let $z=x^{-1}y.$
The other answers look good.
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