Large subgroups of Knuth's non-associative "group" on ${\cal P}(\mathbb{N})$

Question

Donald Knuth introduced a fast, bit-wise approximation to integer addition by $$(a,b) \mapsto a \, ^{\land} \, b \, ^{\land} \, ((a \text{ & } b) \ll 1)$$ where $a,b$ are given in binary and $\,^{\land}$ denotes bitwise ${\tt XOR}$, then $\text{&}$ is bitwise ${\tt AND}$, and finally $\ll 1$ denotes shifting to the left by $1$ position. (Note that $((a \text{ & } b) \ll 1)$ is used to simulate binary carry propagation.)

This can be extended in a nice way giving rise to a binary operation $\oplus:{\cal P}(\mathbb{N})\times {\cal P}(\mathbb{N}) \to {\cal P}(\mathbb{N})$. For $A \in {\cal P}(\mathbb{N})$, let $A+1 = \{a+1: a\in A\}$, and for $A, B\in {\cal P}(\mathbb{N})$ we let $A \, \triangle \, B = (A \setminus B) \cup (B\setminus A)$ be the symmetric difference. The construction $A \, \triangle \, B$ plays the role of bit-wise ${\tt XOR}$ and $A+1$ simulates the left-shift.

Then for $A,B \in {\cal P}(\mathbb{N})$ we let $$A \oplus B := (A \, \triangle \, B) \, \triangle \, ((A \cap B) + 1).$$

The empty set $\emptyset \in {\cal P}(\mathbb{N})$ is the neutral element, the operation is commutative, but not associative: if $A = \{0\}$ and $C = \{1\}$, then $(A \oplus A) \oplus C = \{2\}$, but $A \oplus (A \oplus C) = \emptyset$.

However, every element of ${\cal P}(\mathbb{N})$ has an inverse with respect to $\oplus$.

Let us call $U\subseteq {\cal P}(\mathbb{N})$ a subgroup of ${\cal P}(\mathbb{N})$ if it is closed under $\oplus$ and taking inverses, and if $U$ is associative. Zorn's Lemma implies that every subgroup is contained in a maximal subgroup of ${\cal P}(\mathbb{N})$ with respect to set inclusion $\subseteq$.

Question. Do all maximal subgroups of ${\cal P}(\mathbb{N})$ have the same cardinality? And is there an uncountable subgroup of ${\cal P}(\mathbb{N})$?

(Only one of the questions needs to be fully answered for acceptance.)

Answer 1

Following the suggestion in a comment by Achim Krause I will show that the only solution of the equation $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$ in the magma $(\mathcal P(\mathbb N),\oplus)$ is $x=\emptyset$, whence $\{\emptyset\}$ is the only associative submagma of $(\mathcal P(\mathbb N),\oplus)$.

It will be convenient to identify an element of $\mathcal P(\mathbb N)$ with its characteristic function as a subset of $\mathbb Z$, so that for $x\in\mathcal P(\mathbb N)$ and $n\in\mathbb Z$ we have $x_n=1$ if $n\in x$ and $x_n=0$ if $n\notin x$; in particular $x_n=0$ for all $n\lt0$. Thus for all $n\in\mathbb Z$ and $x,y\in\mathcal P(\mathbb N)$ we have $(x\oplus y)_n=x_n+y_n+x_{n-1}y_{n-1}$ (addition modulo $2$).

Now, for any $x\in\mathcal P(\mathbb N)$ and any $n\in\mathbb Z$, we have $$(x\oplus x)_n=x_n+x_n+x_{n-1}x_{n-1}=x_{n-1}$$ and $$((x\oplus x)\oplus(x\oplus x))_n=(x\oplus x)_{n-1}=x_{n-2},$$ so that $$((x\oplus x)\oplus(x\oplus x))_{n+2}=x_n.\tag1$$ Also $$(x\oplus(x\oplus x))_n=x_n+x_{n-1}+x_{n-1}x_{n-2}$$ and $$(x\oplus(x\oplus(x\oplus x)))_n=$$$$x_n+(x_n+x_{n-1}+x_{n-1}x_{n-2})+x_{n-1}(x_{n-1}+x_{n-2}+x_{n-2}x_{n-3})$$$$=x_{n-1}x_{n-2}x_{n-3},$$ so that $$(x\oplus(x\oplus(x\oplus x)))_{n+2}=x_{n+1}x_nx_{n-1}.\tag2$$ If $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$, then from $(1)$ and $(2)$ we have $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=$$$$(x\oplus(x\oplus(x\oplus x)))_{n+2}=x_{n+1}x_nx_{n-1},$$ whence $x_n=1\implies x_{n+1}=x_{n-1}=1$, i.e., $x$ is a constant function. Since $x_n=0$ for $n\lt0$, it follows that $x_n=0$ for all $n$.