Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 2 / 3

For $|G|=p^{a}m$ with $p\nmid m$, Sylow theorems assert existence of a subgroup of order $p^a$, containment of every p-subgroup in one, conjugacy of all these Sylow subgroups, andA group of order $20$ has $n_5\mid 4$ and $n_5\equiv 1(\mathrm{mod}5)$, so its Sylow 5-subgroup $P$ is unique. In $S_6$ it is generated by a five-cycle and has one fixed point. Its normalizer preserves that point and, on the other five points identified with ${\mathbb{F}}_5$, is the affine group $x\mapsto ax+b$. Every automorphism of the cyclic translation group is realized by a multiplier. Thus $|N_{S_6}(P)|=5\cdot 4=20$, forcing

Inside $S_5$ there are $120/20=6$ Sylow 5-subgroups. Conjugation gives a transitive action on them and a homomorphism into $S_6$. Its group homomorphism kernel is a normal subgroup of $S_5$, whose only possibilities are $1,A_5,S_5$. The latter two would leave an image of order at most two, incompatible with transitivity on six points. Hence it is faithful. Denote the resulting transitive copy of $S_5$ by $J\le S_6$.

The action of $S_6$ on the six cosets of $J$ is also faithful: its group homomorphism kernel cannot contain the simple subgroup $A_6$, since an image of order at most two is not transitive of degree six, and a normal subgroup disjoint from $A_6$ centralizes $A_6$ and is trivial. It therefore gives an automorphism $\tau :S_6\to S_6$ carrying $J$ to a natural point stabilizer. This automorphism is outer, because conjugation cannot carry a transitive subgroup to an intransitive one.

There are precisely two classes of copies of $S_5$ in $S_6$. A faithful intransitive copy must have an orbit of length five: all partitions of six into smaller orbits embed it in groups of order below $120$. Thus it is a natural point stabilizer. A transitive copy is a degree-six coset action of an abstract $S_5$, with an order-$20$ stabilizer. Every order-$20$ subgroup of $S_5$ is a Sylow-five normalizer, so these stabilizers are conjugate and the degree-six actions are permutation-equivalent. Hence all transitive copies form one class. The outer automorphism constructed above swaps these two classes; it cannot preserve the point-stabilizer class, since any automorphism preserving that class induces a relabeling of its six members, and equivariance of their faithful conjugation action then makes the automorphism inner.

Identify $S_6$ with its inner automorphism group inside $\overline{G}$. Every Sylow 5-subgroup of $\overline{G}$ is contained in $S_6$, since that normal subgroup has index two. There are $720/20=36$ of them. ConsequentlyTo prove maximality, first classify the proper overgroups of $N$ in $S_6$. The affine multiplier in $N$ of order four is a four-cycle and is odd. Within $H\cong S_5$, a proper overgroup larger than $N$ could only have order $60$, but the index-two subgroup is $A_5$ and cannot contain that odd element. Thus $N$ is maximal in $H$. A nontransitive overgroup of $N$ must fix its unique fixed point, so is $H$ if it is larger than $N$. A transitive overgroup $M$ has $M\cap H\ge N$; this point stabilizer is either $N$ or $H$. In the latter case $M=S_6$, and in the former $|M|=6\cdot 20=120$, making it a transitive copy of $S_5$.

There is exactly one such transitive copy containing $N$. There are six conjugates of $J$: applying the preceding overgroup argument to its own Sylow-five normalizer shows that $J$ is maximal, and $J$ is not normal. Each contains six Sylow 5-subgroups. Counting pairs of a conjugate copy and a Sylow subgroup gives $6\cdot 6=36$, with the same number through every one of the 36 Sylow subgroups of $S_6$. Thus each Sylow subgroup, and hence its normalizer $N$, is contained in exactly one transitive copy. In particular the only proper strict overgroups of $N$ are its natural $H$ and this $J$.

The subgroup $\overline{N}$ has an outer element, because it has twice the order of $N$. Such an element normalizes $P$ and swaps the unique overgroups $H,J$. If $\overline{N}\le U\le \overline{G}$, then $U\cap S_6$ is normalized by $\overline{N}$, so it cannot be just $H$ or just $J$. It is therefore $N$ or $S_6$. In the latter case $U=\overline{G}$. In the former $U$ normalizes $N$ and hence its unique Sylow 5-subgroup $P$, giving $U\le \overline{N}$. Therefore