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.

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.
