= Solution
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>, and
$$
n_p\equiv1\pmod p,\qquad n_p\mid m,\qquad n_p=[G:N_G(P)].
$$
A <group> of order $20$ has $n_5\mid4$ and $n_5\equiv1\pmod5$, 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 $|N_{S_6}(P)|=5\cdot4=20$, forcing
$$
\boxed{N=N_{S_6}(P)\cong C_5\rtimes C_4,\qquad N\le H\cong S_5.}
$$
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. Consequently
$$
\boxed{|\overline N|=1440/36=40,\qquad\overline N\cap S_6=N.}
$$
To 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 $|M|=6\cdot20=120$, 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 $6\cdot6=36$, 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
$$
\boxed{\overline N\text{ is a maximal subgroup of }\operatorname{Aut}(S_6).}
$$
Back to article page
