# Orbital maximin equals size of set for abelian groups

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## Statement

Suppose $n$ is a natural number greater than $1$. Then, for any abelian group $G$ acting on a set of size $n$, the maximum possible value of the minimum of the sizes of the orbitals is $n$. Further, there exists an abelian group (namely, the cyclic group of order $n$) for which this maximum is achieved.

This is a resolution to the Orbital maximin problem (?) for abelian groups.

## Proof

We assume that $n$ is at least $3$. The result holds trivially for $n = 2$.

### Reducing to the case of a transitive action

Suppose the action of $G$ on a set $S$ of size $n$ has more than one orbit. Let $m$ be the size of the smallest orbit, which we call $\mathcal{O}$. $G$ restricts to a permutation action on $\mathcal{O}$. In particular, the orbit of any orbital with both points in $\mathcal{O}$ lies completely inside the orbital set for $\mathcal{O}$.

Thus, if we prove the result for all $m$ for a transitive action, then using $m \le n$ would complete the proof. We can thus assume that the action is transitive.

### Every element must act freely (semiregularly), i.e., without fixed points

We assume $G$ acts transitively on a set $S$ of size $n$.

Suppose we have an element $\sigma$ in $G$ and elements $x,y,z \in S$ such that $\sigma(x) = x$ and $\sigma(y) = z$. Since $G$ acts transitively, there exists $\tau$ such that $\tau(x) = y$. Then, $\tau(\sigma(x)) = \tau(x) = y$ whereas $\sigma(\tau(x)) = \sigma(y) = z$. Thus, $\sigma \circ \tau \ne \tau \circ \sigma$, contradicting abelianness.

Note that we can in fact deduce that the action must be a regular group action since it is both transitive and semiregular. This is part of a more general idea: any action that centralizes a transitive group action must be semiregular.

### The size of an orbital is exactly $n$ for a transitive action

As before, $G$ acts transitively on a set $S$ of size $n$.

Consider any pair $(x,y), x \ne y$. We know that for any two elements $\sigma_1, \sigma_2$, if $\sigma_1(x) = \sigma_2(x)$, then $\sigma_2^{-1}(\sigma_1(x)) = x$. Since the action is free, this forces $\sigma_1 = \sigma_2$, also forcing $\sigma_1(y) = \sigma_2(y)$. Thus, no two distinct elements of the orbital of $(x,y)$ can have the same first coordinate. Since the number of first coordinates is $n$, we obtain that the size of the orbital is exactly $n$.