Let $X$ be an infinite set, and $k \gg 1$ be a natural number. We work without the axiom of choice.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-times power set of $X$) in a natural way, with $G_0$ a subgroup of $G_1$. So in ${\cal P}^k(X)$ the orbits of $G_1$ split into orbits of $G_0$. Let $a$ and $b$ be two elements of ${\cal P}^k(X)$ that belong to the same $G_1$ orbit.

Is there a $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ (which both belong to the same $G_1$-orbit as $a$ and $b$) nevertheless belong to different $G_0$ orbits?

I am looking for a result that: independent of the axiom of choice, for all infinite $X$, for all sufficiently large $k$, and for all distinct $a$ and $b$ belonging to the same $G_1$-orbit, there is $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ belong to different $G_0$ orbits.

A positive result would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!

Let $X$ be an infinite set, and $k \ge 1$ be a natural number. We work without the axiom of choice.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-fold power set $\mathcal{P}(\mathcal{P}(\cdots(X)\cdots)$ of $X$) in a natural way, with $G_0$ a subgroup of $G_1$. So in ${\cal P}^k(X)$ the orbits of $G_1$ split into orbits of $G_0$.

Is the following true in ZF: for all $k\ge 1$, for all distinct elements $a\neq b$ of ${\cal P}^k(X)$ that belong to the same $G_1$-orbit, there exists $\sigma \in G_1$ such that $\sigma(a)$ and $\sigma(b)$ belong to distinct $G_0$-orbits?

A positive result would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!

Let $X$ be an infinite set, and $k \gg 1$ be a natural number. We work without the axiom of choice.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-times power set of $X$) in a natural way. $G_0$ can be thought of as a subgroup of $G_1$ in a natural way, so it is natural to think of orbits of $G_1$ acting on ${\cal P}^k(X)$ as splitting into orbits of $G_0$ acting on ${\cal P}^k(X)$. Let $a$ and $b$ be two elements of ${\cal P}^k(X)$ that belong to the same $G_1$ orbit.

Is there a $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ (which both belong to the same $G_1$-orbit as $a$ and $b$) nevertheless belong to different $G_0$ orbits?

I am looking for a result that: independent of the axiom of choice, for all infinite $X$, for all sufficiently large $k$, and for all distinct $a$ and $b$ belonging to the same $G_1$-orbit, there is $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ belong to different $G_0$ orbits.

A `yes' would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!

Let $X$ be an infinite set, and $k \gg 1$ be a natural number. We work without the axiom of choice.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-times power set of $X$) in a natural way, with $G_0$ a subgroup of $G_1$. So in ${\cal P}^k(X)$ the orbits of $G_1$ split into orbits of $G_0$. Let $a$ and $b$ be two elements of ${\cal P}^k(X)$ that belong to the same $G_1$ orbit.

Is there a $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ (which both belong to the same $G_1$-orbit as $a$ and $b$) nevertheless belong to different $G_0$ orbits?

I am looking for a result that: independent of the axiom of choice, for all infinite $X$, for all sufficiently large $k$, and for all distinct $a$ and $b$ belonging to the same $G_1$-orbit, there is $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ belong to different $G_0$ orbits.

A positive result would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!

Head coming up above parapet .. NOW! I have noticed that as soon as i ask a question publicly the answer becomes trivially obvious, and i end up feeling an idiot. However, this one has vexed me for some time, and i've tried it on better men than me and they haven't cracked it. It may be that i haven't explained it properly, but i t-h-i-n-k that the text which follows explains it properly...

Let $X$ be an infinite set, and $k >> 1$ be a natural number. Put all thought of the axiom of choice out of your mind. And try to answer the following question without making any special assumptions about $k$ or $X$.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-times power set of $X$) in a natural way. $G_0$ can be thought of as a subgroup of $G_1$ in a natural way, so it is natural to think of orbits of $G_1$ acting on ${\cal P}^k(X)$ as fissioning into orbits of $G_0$ acting on ${\cal P}^k(X)$. Let $a$ and $b$ be two elements of ${\cal P}^k(X)$ that belong to the same $G_1$ orbit. Clearly we cannot expect them to belong to two different $G_0$ orbits, but we might be able to arrange for the following. Is there a $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ (which both, after all, belong to the same $G_1$-orbit as $a$ and $b$) nevertheless belong to different $G_0$ orbits?

The result i am looking for is a result that says that: for all infinite $X$, for all sufficiently large $k$, and for all distinct $a$ and $b$ belonging to the same $G_1$-orbit, there is $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ belong to different $G_0$ orbits.

The answer really ought to be `yes' and it would have nice consequences for the model theory of Russell-Ramsey typed set theory, but i'll say nothing about that for the moment!

Let $X$ be an infinite set, and $k \gg 1$ be a natural number. We work without the axiom of choice.

Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the full symmetric group on ${\cal P}(X)$, the power set of $X$. Both these groups act on ${\cal P}^k(X)$ (the $k$-times power set of $X$) in a natural way. $G_0$ can be thought of as a subgroup of $G_1$ in a natural way, so it is natural to think of orbits of $G_1$ acting on ${\cal P}^k(X)$ as splitting into orbits of $G_0$ acting on ${\cal P}^k(X)$. Let $a$ and $b$ be two elements of ${\cal P}^k(X)$ that belong to the same $G_1$ orbit.

Is there a $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ (which both belong to the same $G_1$-orbit as $a$ and $b$) nevertheless belong to different $G_0$ orbits?

I am looking for a result that: independent of the axiom of choice, for all infinite $X$, for all sufficiently large $k$, and for all distinct $a$ and $b$ belonging to the same $G_1$-orbit, there is $\sigma \in G_1$ s.t. $\sigma(a)$ and $\sigma(b)$ belong to different $G_0$ orbits.

A `yes' would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!