Ring Theory question in a GRE practice exam I have a question about a GRE practice problem relating to Rings. The question is as follows:
Suppose that two binary operations, denoted by $\oplus$ and $\odot$ , are defined on a nonempty set $S$, and that the
following conditions are satisfied for all $x, y$, and $z$ in $S$:
(1) $(x\oplus y)$ and $x \odot y$ are in S.
(2) $x\oplus (y \oplus z) =  (x \oplus y) \oplus z$ and $x \odot (y \odot z) = (x \odot y) \odot z$.
(3) $x \oplus y = y \oplus x$
Also, for each $x$ in $S$ and for each positive integer $n$, the elements $nx$ and $xn$ are defined recursively as
follows:
$1x =  x^1 = x$ and
if $kx$ and $x^k$ have been defined, then $(k + 1) x = kx \oplus x$ and $x^{k+1} =  x^k \odot x$.
Which of the following must be true?
I) $(x \odot y)^n = x^n \odot y^n$ for all $x,y \in S$ and all positive integers $n$.
II) $n(x \oplus y) = nx \oplus ny$ for all $x,y \in S$ and all positive integers $n$.
III) $x^m \odot x^n = x^{m+n}$ for all $x \in S$ and all positive integers $n,m$.
A) I only, B) II only, C) III only, D) II and III, E) I, II and III
I recognize that the description given is the description of a ring. So I was pretty sure that the answer was E) I, II, III. It is pretty easy to show II and III, and I couldn't think of a counterexample for I. However, the solutions manual I purchased says the answer is D) II and III, claiming that the ring of integers does not have property I.
So my question is, what is the darn counterexample? I can't think of any. I tried $x=1$ and $y=-1$ with both odd and even powers, and even tried $x=1$, $y=2$ and $x=-1$, $y=2$, still nothing. Any suggestions? Thanks in advance.

Here's a counterexample for (I). $\mathrm{Mat}_{2 \times 2}(\mathbb{R})$ equipped with matrix addition and multiplication satisfies all of the above criteria, and we consider the matrices $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \: \: B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$
For $n = 2$, we have that $$(AB)^2 = \left(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\right)^2 = \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ 
but $$A^2 B^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$$ 
So, $A^2B^2 \ne (AB)^2$, and therefore this is a counterexample for (I).
