# 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.

• I am stunned to see such a question in a GRE – imranfat Sep 11 '14 at 21:03
• You do not necessarily have a ring. In particular there are no identity elements, you have no additive inverses, and you aren't told that multiplication distributes over addition. 1 is going to be false because you aren't told that multiplication is commutative. – James Sep 11 '14 at 21:09
• It's not a ring - there is no hint that there are additive inverses or an identity in either operator. For example, the positive integers with the usual operators satisfies the above. – Thomas Andrews Sep 11 '14 at 21:11
• This question is simpler that it appears at first glance. The key is (3) implies $\odot$ may not be commutative which breaks the induction step when trying to show I – illysial Sep 11 '14 at 21:14
• @CarlMummert Imagine if this question did appear on the math portion of the ordinary GRE...you would hear the collective sound of heads exploding across the country. – heropup Sep 11 '14 at 21:27

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).

Claims II and III are necessarily true from the given conditions. Proof by induction: Suppose $(n-1)x \oplus (n-1)y = (n-1) z$, where $z = x \oplus y$. Then by definition, $$nx \oplus ny = ((n-1)x \oplus x) \oplus ((n-1)y \oplus y).$$ But property (3) indicates that $\oplus$ is commutative, so the above can be arranged at will: $$nx \oplus ny = ((n-1) x \oplus (n-1) y) \oplus (x \oplus y) = (n-1)z \oplus z = nz.$$ Claim III is an obvious and direct consequence of the associativity of $\odot$.

The only claim that is not necessarily true is I, because this requires $\odot$ to be commutative, which is not postulated. This also suggests a choice of set $S$ for which properties (1-3) hold true, yet claim I is false; e.g., try a set of $2 \times 2$ matrices.

Your solution manual is incorrect that the integers are a counterexample, because the integers are a commutative ring.

For a true counterexample to I, take the $2\times 2$ matrices over the integers with addition as matrix addition and multiplication as matrix multiplication, with $x = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$ and $y = \begin{pmatrix} 1 & 2 \\ 1 & 1\end{pmatrix}$.

The questioner is trying to deceive you. What you have are two independent operations on the set $S$; both are associative, but only one is commutative.

Under either operation, $S$ is a semigroup. Let's use the common notation for semigroups, that is, simple juxtaposition denotes the result of the operation; associativity is expressed by $$x(yz)=(xy)z \quad\text{for all x,y,z}$$ Powers are defined recursively by $$x^1=x,\quad x^{n+1}=x^nx.$$ Proving that $$x^{m+n}=x^mx^n$$ is standard: it's true by definition when $n=1$; suppose it holds for $n$: then \begin{align} x^{m+n+1}&=x^{m+n}&&\text{definition}\\ &=(x^mx^n)x&&\text{induction hypothesis}\\ &=x^m(x^nx)&&\text{associativity}\\ &=x^mx^{n+1}&&\text{definition} \end{align}

This proves III.

If the semigroup is commutative, that is $xy=yx$ for all $x,y$, then also $$(xy)^n=x^ny^n$$ holds. This can be proved by a double induction: first show that $y^nx=xy^n$ and then $(xy)^n=x^ny^n$. This proves II, because placing the exponent on the left instead as a superscript is just a matter of notation.

For a counterexample to I, consider the group $S_3$ under composition; let $x$ be the permutation that swaps $2$ and $3$ and $y$ the permutation that swaps $1$ and $2$. Then $(xy)^2$ is not the identity, while $x^2y^2$ is the identity. This can be the $\odot$ operation; for $\oplus$ consider the operation that composes any two objects producing the identity: this is associative and commutative.