Is this function of partitions one-to-one? Suppose we have a set of integers $H=\{1,2, ...n\}$. Let $A$ a set of partitions of H into $n/2$ pairs $\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\}$ and
function $f:A \rightarrow Z^n$ where $f(\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\})=\cup \{x_i*y_i\}$.
For a set of integers $\{1,2, ...8\}$ and the pairing $\{\{3,6\},\{1,7\},\{2,4\},\{5,8\}\}$
$f(\{\{3,6\},\{1,7\},\{2,4\},\{5,8\}\})=\{18,7,8,40\}$

Is this function injective?

 A: Here are two partitions of $\{1,2,3,4,5,6,7,8\}$ that give the same products:
$\{\{1,6\}, \{2,4\}, \{3,8\}, \{5,7\}\}$ and $\{\{2,3\},\{1,8\},\{4,6\},\{5,7\}\}$
It is perhaps worth observing that any prime in the upper half of the interval must pair with the same thing in both partitions.
A: No.  For $n$ large enough, $f$ will not be injective.
It's easier to consider the case of addition, where we have
$$1 + (-1) = 0, \qquad 2 + (-3) = -1, \qquad 3 + (-2) = 1 \\
1 + (-2) = -1, \qquad 2 + (-1) = 1, \qquad 3 + (-3) = 0$$
That is, partition $\{-3,-2,-1, 1, 2, 3\}$ into three parts of size $2$ in at least two ways that give the same set of sums.
This answers your question by applying the map $x \mapsto 2^{x+3}$ to find
$$2^{1+3}2^{(-1)+3} = 2^{0+6}, \qquad 2^{2+3}2^{(-3)+3} = 2^{-1+6}, \qquad 2^{3+3}2^{(-2)+3} = 2^{1+6} \\
2^{1+3}2^{(-2)+3} = 2^{-1+6}, \qquad 2^{2+3}2^{(-1)+3} = 2^{1+6}, \qquad 2^{3+3}2^{(-3)+3} = 2^{0+6}$$
so we find that $f$ is not injective for $n \geq 2^6 =64$. [it likely fails to be injective for smaller values as well]
