Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 9E (Groups, Rings and Modules)

Let $Q$ act by left multiplication on the set $G/P$ of left cosets. Every orbit has size a power of $p$, whileis not divisible by $p$. Therefore at least one orbit has size one. If $gP$ is fixed, then $qgP=gP$ for every $q\in Q$, so $g^{-1}Qg\le P$, equivalentlyThis is Sylow containment from a coset fixed point.

The remaining Sylow theorems state that Sylow $p$-subgroups exist, that any two are conjugate, and that their number $n_p$ satisfies

By transitivity and the orbit-stabilizer theorem,Inside $G_x\cong S_4$, there are three Sylow $2$-subgroups, each of order eight, and four Sylow $3$-subgroups, each of order three. These are also Sylow subgroups of $G$.

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

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

The Sylow congruence and divisibility conditions give $n_7\in \{1,8\}$. Regard the faithful action as an embedding $G\le S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\le N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot 6=42$, impossible for a subgroup $G$ of order 168. HenceThese are the Sylow counts in a faithful degree-seven action with S4 point stabilizers.

Let $N⊴G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem givesSince $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal subgroup is transitive by normal subgroup orbits in a faithful prime-degree action. HenceBut $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal subgroup has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such subgroup would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal subgroup, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal subgroup. Therefore $G$ is simple.