Return to Answer

Embarrassingly, the posting below contains another incomplete proof, as was pointed out by Hugo Chapdelaine. I am working on a new (third) version, but after two wrong proofs, I want to hold off for a while on posting it until it is written up carefully and checked. What I am trying to prove is this:

THEOREM: Let $H$ be a transitive subgroup of $S_N$, and suppose $M \triangleleft H$ and $H/M \cong S_n$, where $N/2 \le n < N$ and $5 < n$. Then $n = N/2$ and all orbits of $M$ have size $2$.

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)

Embarrassingly, the posting below contains another incomplete proof, as was pointed out by Hugo Chapdelaine. I am working on a new (third) version, but after two wrong proofs, I want to hold off for a while on posting it until it is written up carefully and checked. What I am trying to prove is this:

THEOREM: Let $H$ be a transitive subgroup of $S_N$, and suppose $M \triangleleft H$ and $H/M \cong S_n$, where $N/2 \le n < N$ and $5 < n$. Then $n = N/2$ and either $M = 1$ or all orbits of $M$ have size $2$.

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)

Embarrassingly, the posting below contains another incomplete proof, as was pointed out by Hugo Chapdelaine. I am working on a new (third) version, but after two wrong proofs, I want to hold off for a while on posting it until it is written up carefully and checked. What I am trying to prove is this:

THEOREM: Let $H$ be a transitive subgroup of $S_N$, and suppose $M \triangleleft H$ and $H/M \cong S_n$, where $N/2 \le n < N$ and $5 < n$. Then $n = N/2$ and all orbits of $M$ have size $2$.

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ an $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)

Unfortunately, the proof in the original version of this post had a gap that I do not see how to repair. The following weaker result seems to be true, however.

THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and $M \cap K = 1$. In particular, $K \cong S_n$.

Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write $u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of $H/M \cong S_n$, so this index is either $1$ or at least $n$.

Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so $(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and $H = M(H \cap S) = M(M \cap S)K = MK$ as required.

We can assume now that for every choice of point stabilizer $S$ we have $u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED

In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if $n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)