= Solution
Count pairs $(x,P)$ where $P$ is a Sylow $2$-subgroup fixing $x$. Each of the seven point stabilizers contains three such <subgroups>, giving 21 pairs. Conversely, a $2$-group acting on seven points has a fixed point, and a Sylow $2$-subgroup cannot fix two points because a two-point stabilizer has order four. Hence every Sylow $2$-subgroup occurs in exactly one pair, so
$$
n_2=21.
$$
Similarly, each point stabilizer contains four Sylow $3$-subgroups, giving 28 pairs. A <group> of order three acting on seven points has a fixed point, and it cannot fix two because a two-point stabilizer is a $2$-group. Thus
$$
n_3=28.
$$
The Sylow congruence and divisibility conditions give $n_7\in\{1,8\}$. Regard the faithful action as an embedding $G\leq S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\leq N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot6=42$, impossible for a <subgroup> $G$ of order 168. Hence
$$
n_7=8.
$$
These are the <Sylow counts in a faithful degree-seven action with S4 point stabilizers>.
Back to article page
