Putnam 2010, Problem A-5

Let \( G \) be a group, with operation \( * \). Suppose that
(i) \( G \) is a subset of \( \mathbb{R}^3 \) (but \( * \) need not be related to addition of vectors);
(ii) For each \( \mathbf{a}, \mathbf{b} \in G \), either \( \mathbf{a} \times \mathbf{b} = \mathbf{a} * \mathbf{b} \) or \( \mathbf{a} \times \mathbf{b} = 0 \) (or both), where \( \times \) is the usual cross product in \( \mathbb{R}^3 \).

Prove that \( \mathbf{a} \times \mathbf{b} = 0 \) for all \( \mathbf{a}, \mathbf{b} \in G \).

Solution:
We start with three lemmas.

**Lemma 1.** If $\mathbf{x}, \mathbf{y} \in G$ are nonzero orthogonal vectors, then $\mathbf{x} * \mathbf{x}$ is parallel to $\mathbf{y}$.

*Proof.* Put $\mathbf{z} = \mathbf{x} \times \mathbf{y} \neq 0$, so that $\mathbf{x}, \mathbf{y}$, and $\mathbf{z} = \mathbf{x} * \mathbf{y}$ are nonzero and mutually orthogonal. Then $\mathbf{w} = \mathbf{x} \times \mathbf{z} \neq 0$, so $\mathbf{w} = \mathbf{x} * \mathbf{z}$ is nonzero and orthogonal to $\mathbf{x}$ and $\mathbf{z}$. However, if $(\mathbf{x} * \mathbf{x}) \times \mathbf{y} \neq 0$, then $\mathbf{w} = \mathbf{x} * (\mathbf{x} * \mathbf{y}) = (\mathbf{x} * \mathbf{x}) * \mathbf{y} = (\mathbf{x} * \mathbf{x}) \times \mathbf{y}$ is also orthogonal to $\mathbf{y}$, a contradiction. $\square$

**Lemma 2.** If $\mathbf{x} \in G$ is nonzero, and there exists $\mathbf{y} \in G$ nonzero and orthogonal to $\mathbf{x}$, then $\mathbf{x} * \mathbf{x} = 0$.

*Proof.* Lemma 1 implies that $\mathbf{x} * \mathbf{x}$ is parallel to both $\mathbf{y}$ and $\mathbf{x} \times \mathbf{y}$, so it must be zero. $\square$

**Lemma 3.** If $\mathbf{x}, \mathbf{y} \in G$ commute, then $\mathbf{x} \times \mathbf{y} = 0$.

*Proof.* If $\mathbf{x} \times \mathbf{y} \neq 0$, then $\mathbf{y} \times \mathbf{x}$ is nonzero and distinct from $\mathbf{x} \times \mathbf{y}$. Consequently, $\mathbf{x} * \mathbf{y} = \mathbf{x} \times \mathbf{y}$ and $\mathbf{y} * \mathbf{x} = \mathbf{y} \times \mathbf{x} \neq \mathbf{x} * \mathbf{y}$. $\square$

We proceed now to the proof. Assume by way of contradiction that there exist $\mathbf{a}, \mathbf{b} \in G$ with $\mathbf{a} \times \mathbf{b} \neq 0$. Put $\mathbf{c} = \mathbf{a} \times \mathbf{b} = \mathbf{a} * \mathbf{b}$, so that $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are nonzero and linearly independent. Let $\mathbf{e}$ be the identity element of $G$. Since $\mathbf{e}$ commutes with $\mathbf{a}, \mathbf{b}, \mathbf{c}$, by Lemma 3 we have $\mathbf{e} \times \mathbf{a} = \mathbf{e} \times \mathbf{b} = \mathbf{e} \times \mathbf{c} = 0$. Since $\mathbf{a}, \mathbf{b}, \mathbf{c}$ span $\mathbb{R}^3$, $\mathbf{e} \times \mathbf{x} = 0$ for all $\mathbf{x} \in \mathbb{R}^3$, so $\mathbf{e} = 0$.

Since $\mathbf{b}, \mathbf{c}$, and $\mathbf{b} \times \mathbf{c} = \mathbf{b} * \mathbf{c}$ are nonzero and mutually orthogonal, Lemma 2 implies
$$\mathbf{b} * \mathbf{b} = \mathbf{c} * \mathbf{c} = (\mathbf{b} * \mathbf{c}) * (\mathbf{b} * \mathbf{c}) = 0 = \mathbf{e}.$$

Hence $\mathbf{b} * \mathbf{c} = \mathbf{c} * \mathbf{b}$, contradicting Lemma 3 because $\mathbf{b} \times \mathbf{c} \neq 0$. The desired result follows.