IMC 2007 · Day 1 · P4

We start with three preliminary observations.

Let $U, V$ be two arbitrary subsets of $G$. For each $x \in U$ and $y \in V$ there is a unique $z \in G$ for which $xyz = e$. Therefore, $$\tag{1} N_{UVG} = |U \times V| = |U| \cdot |V|.$$ Second, the equation $xyz = e$ is equivalent to $yzx = e$ and $zxy = e$. For arbitrary sets $U, V, W \subset G$, this implies $$\{ (x,y,z) \in U \times V \times W : xyz = e \} = \{ (x,y,z) \in U \times V \times W : yzx = e \} = \{ (x,y,z) \in U \times V \times W : zxy = e \}$$ and therefore $$\tag{2} N_{UVW} = N_{VWU} = N_{WUV}.$$ Third, if $U, V \subset G$ and $W_1, W_2, W_3$ are disjoint sets and $W = W_1 \cup W_2 \cup W_3$ then, for arbitrary $U, V \subset G$, $$\begin{align*} \{ (x,y,z) \in U \times V \times W : xyz = e \} &= \{ (x,y,z) \in U \times V \times W_1 : xyz = e \} \cup \\ &\quad \cup \{ (x,y,z) \in U \times V \times W_2 : xyz = e \} \cup \{ (x,y,z) \in U \times V \times W_3 : xyz = e \} \end{align*}$$ so $$\tag{3} N_{UVW} = N_{UVW_1} + N_{UVW_2} + N_{UVW_3}.$$ Applying these observations, the statement follows as $$\begin{align*} N_{ABC} &= N_{ABG} - N_{ABA} - N_{ABB} = |A| \cdot |B| - N_{BAA} - N_{BAB} = \\ &= N_{BAG} - N_{BAA} - N_{BAB} = N_{BAC} = N_{CBA}. \end{align*}$$