TITLE: When does $G\times G\times G$ admit a faithful group action on a set of size $|G|$?
QUESTION [22 upvotes]: [Edited due to YCor's comment:]
Given a finite group $G$, under what conditions does $G\times G\times G$ (the direct product of three copies of $G$) admit a faithful group action on a set of size $|G|$? This is equivalent to an injective group homomorphism from $G\times G\times G$ to $S_{|G|}$.
For reference, for every group $G$ with a trivial center, $G\times G$ admits an easy faithful group action on $G$ by $(g_1, g_2): g \rightarrow g_1 g g_2^{-1}$.

REPLY [8 votes]: This is more of an extended comment than an answer. I will determine all the abelian groups failing to act faithfully on at most $n/3$ points.
Suppose $G$ is abelian, say $G = C_{q_1} \times C_{q_2} \times \cdots \times C_{q_k}$, where $q_1, q_2, \dots$ are (not necessarily distinct) prime powers. Then the minimal faithful permutation action of $G$ has $q_1 + q_2 + \cdots + q_k$ points. For a proof see https://mathoverflow.net/a/409831/20598.
Hence $G$ does not act faithully on $|G|/3$ points iff
$$q_1 + \cdots + q_k > q_1 \cdots q_k / 3.$$
It is maybe easier to think of this as
$$ \sum_{i=1}^k q_i / (q_1 \cdots q_k) > 1/3 . $$
Each of the terms here is at most $1/2^{k-1}$, so $k / 2^{k-1} > 1/3$, so $k \leq 4$.
By further analyzing $k=1,2,3,4$ we find all the solutions (where $q_1 \leq \cdots \leq q_k$):
$$
(q_1),\\
(2, q_2),
(3, q_2),\\
(4, 4), (4, 5), (4, 7), (4, 8), (4, 9), (4, 11),
(5, 5), (5, 7),\\
(2, 2, q_3) \qquad (2 \leq q_3 \leq 11),\\
(2, 3, 3), (2, 3, 4),
(2, 2, 2, 2), (2, 2, 2, 3).
$$