Forming a group from power set pairs

Question

Question I want to form a group out of disjoint pairs taken from the power set $\mathcal{P}(X)$ over a finite set $X$. An element of the group $G$ is a pair $u=(u_1,u_2) \in \mathcal{P} \times \mathcal{P}$ such that $u_1 \cap u_2 = \emptyset$. Are there any ways to form a group on these disjoint pairs of elements?

Attempt I thought to use the symmetric difference $a \Delta b = (a \cup b) \backslash (a \cap b)$ as done for the regular power set group, but for both pairs. Take $u,v \in G$ and the group operation gives $w$ defined as: \begin{eqnarray} w = u \circ v &=& (u_1 \Delta v_1,\;u_2 \Delta v_2 \backslash [(u_1 \Delta v_1) \cap (u_2 \Delta v_2)])\\ &=&(u_1 \Delta v_1,\;(u_2 \Delta v_2) \backslash (u_1 \Delta v_1)). \end{eqnarray} The term in the square brackets ensures that the new pair is disjoint: $w_1 \cap w_2 = \emptyset$. This is closed in $G$ since $G$ contains every pair of disjoint sets. To form a group: (a) this has an identity $(\emptyset, \emptyset)$, (b) each element is its own inverse (an involution), (c) I'm unclear about associativity. I tried showing that it's associative by expanding both $a \circ (b \circ c)$ and $(a \circ b) \circ c$, but I'm running into some trouble. So far, I have: \begin{eqnarray} a \circ (b \circ c) &=& a \circ (b_1 \Delta c_1,\; b_2 \Delta c_2 \backslash [(b_1 \Delta c_1) \cap ( b_2 \Delta c_2)])\\ &=& (a_1 \Delta b_1 \Delta c_1,\; a_2 \Delta (b_2 \Delta c_2 \backslash [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)])\\ \end{eqnarray} and \begin{eqnarray} (a \circ b) \circ c &=& (a_1 \Delta b_1,\; a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)]) \circ c\\ &=& (a_1 \Delta b_1 \Delta c_1,\; (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2)\\ \end{eqnarray} The left parts of the pairs match, but the right parts need some work. So we want to check whether the right parts are equal, i.e. that: $$ a_2 \Delta (b_2 \Delta c_2 \backslash [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)] = (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2 $$ I'm getting stuck here just because this expands (in $\Delta$) to a lot of terms. Perhaps there's a smart to check this?

Answer 1

Your construction cannot give a group if your set has more than one element.

Consider $a=(\{1,2\},\varnothing)$, $b=(\varnothing,\{2\})$, and $c=(\varnothing,\varnothing)$. Then $b\neq c$, but $$\begin{align*} a\circ b&= (\{1,2\},\varnothing)\circ (\varnothing,\{2\})\\ &= (\{1,2\},\{2\}-\{1,2\}) = (\{1,2\},\varnothing).\\ a\circ c&= (\{1,2\},\varnothing)\circ (\varnothing,\varnothing) = (\{1,2\},\varnothing). \end{align*}$$ So $b\neq c$, but $a\circ b= a\circ c$. This is impossible in a group.

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In fact, one element suffices: set $X=\{\star\}$. Then $(X,\varnothing)\circ(\varnothing,X)=(X,\varnothing)=(X,\varnothing)\circ(\varnothing,\varnothing)$, so again you cannot have a group.

(Of course, if $X$ is empty then you have the one-element group...)

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You ask in comments whether you have a loop. The answer is no, unless $X$ is empty. Consider the elements $(X,\varnothing)$ and $(\varnothing,X)$. There is no element $A$ such that $(X,\varnothing)\circ A = (\varnothing, X)$. Indeed, such an element $A=(R,S)$ would necessarily have $X\Delta R=\varnothing$, hence $R=X$; but then the condition $R\cap S=\varnothing$ requires $S=\varnothing$, and $(X,\varnothing)\circ(X,\varnothing)=(\varnothing,\varnothing)\neq (\varnothing, X)$.

But in a loop, the equations $ax=b$ and $ya=b$ have solutions for all $a$ and $b$, so this is not a loop either.

Answer 2

Your operation need not be associative.

Let $X=\{0,1\}$, $a=\langle\{0\},\{1\}\rangle$, and $b=\langle\{1\},\{0\}\rangle$. Then $a\circ b=\langle X,\varnothing\rangle$, so

$$a\circ(a\circ b)=\langle\{0\},\{1\}\rangle\circ\langle X,\varnothing\rangle=\langle\{1\},\varnothing\rangle\,.$$

However, $(a\circ a)\circ b=b\ne\langle\{1\},\varnothing\rangle$.