Subgroups of $S_n$ with all non-trivial elements of same cycle type.

Question

Question: Which subgroups of $S_n$ are maximal with respect to having all their non-trivial elements of the same cycle-type?

Motivation: A recent question asked whether a knowledge of the characters of $S_n$ allows one to deduce the characters of a given subgroup. The commenters pointed out that this was unlikely. It seemed to me that good counter-examples would be subgroups of $S_n$ whose non-trivial elements were all conjugate in $S_n$, since the character table of $S_n$ would not yield any information distinguishing the classes of the subgroup.

Comments:

(i) Clearly any subgroup $H\leqslant S_n$ with all non-trivial elements conjugate in $S_n$ must be a $p$-group of exponent $p$.

(ii) Every $p$-group $H$ of exponent $p$ can be embedded in $S_{|H|}$ via its regular representation. As every non-trivial element yields a fixed-point-free permutation of order $p$ it is a product of $p$-cycles, there being $(|H|/p)$ such cycles. Hence all the non-trivial elements are conjugate in $S_{|H|}$.

(iii) It seems to me that $H$ is a very small subgroup of $S_{|H|}$. Hence my question about the maximal subgroups where all the non-trivial elements are of the same cycle type.

Added after the bounty expired.

(iv) Originally I did not tag this combinatorics, probably wrongly as Joshua P. Swanson points out in his answer/extended commentary. There is a sense in which every question about permutations is a combinatorial question, but Swanson makes clear what sort of combinatorics are involved.

(v) I am still surprised that no one has told me that all this is well-known and/or an easy exercise.

Indeed I am so surprised that I asked ChatGPT to answer the question. Its first attempt was pathetic, assuming for instance that all groups of exponent $p$ are abelian. After putting in that information its second attempt asserted that nevertheless only abelian $p$-groups were embeddable in the required way. However after telling it about regular representations it purported to have a group theoretic proof of the following two theorems. As I can't follow the proofs in detail I won't add them.

Structure Theorem

Let $G \leq S_n $be a nontrivial subgroup with the property that every nonidentity element of $G$ has the same cycle type.

Then there exists a prime $ p $ and a finite group $ H $ of exponent $ p $ such that:

(1) $ G $ is (up to conjugacy in $ S_n $ of the form $$ G = \Delta(H) \leq \operatorname{Sym}(H)^t \leq S_n, $$ where $$ \Delta(H) = \{(L_h, \ldots, L_h) : h \in H\}, $$ and $ L_h $ denotes left translation by $ h $ on $ H $.

(2) The action of $ G $ on the underlying set $ X $ decomposes as $$ X = (H \times \{1, \dots, t\}) \;\sqcup\; F, $$ where $ F $ is a (possibly empty) subset of points fixed by all of $ G $, and on each copy $ H \times \{i\} $ the group $ G $ acts regularly via the left regular representation.

(3) Every nonidentity element of $ G $ has order $ p $ and, on $ X $, is a product of $$ \frac{t \, |H|}{p} $$ disjoint $ p $-cycles together with $ |F| $ fixed points.

Conversely, any group of the above form (for some exponent-$ p $ group $ H $, integer $ t \ge 1 $, and fixed-point set $ F $) has the property that all its nonidentity elements have the same cycle type.

Maximality Criterion

Let $H$ be a finite group of exponent $p$, and let $$ G = \Delta(H) \leq S_X $$ be its diagonal embedding acting regularly on $$ X = (H \times \{1, \dots, t\}) \;\sqcup\; F, $$ with $|F| = f$ fixed points.

Then $G$ is maximal among subgroups of $S_X$ in which every nonidentity element has the same cycle type if and only if both of the following conditions hold:

(1) No index-extension. There is no exponent-$p$ group $K$ with $H < K$ and $[K : H]$ dividing $t$. Equivalently, $H$ is maximal by inclusion among exponent-$p$ groups whose order divides $n - f$.

(2) No compatible automorphism or copy-permutation. There is no nontrivial element $(\alpha, \sigma) \in \operatorname{Aut}(H) \times S_t$ of order $p$ such that the induced permutation $$ \pi(h,i) = (\alpha(h), \sigma(i)) $$ on $H \times \{1, \dots, t\}$ has exactly the same fixed-point pattern as the nonidentity elements of $G$ (that is, order $p$ and precisely $f$ fixed points).

Equivalently, a strict supergroup $L > G$ in $S_X$ with the same uniform cycle-type property exists if and only if at least one of the following holds:

(A) Type A extension. There exists an exponent-$p$ group $K \supsetneq H$ with $[K : H] = k > 1$ dividing $t$, and $L$ is obtained by grouping the $t$ copies of $H$ into $t/k$ blocks of $k$ copies each and replacing each diagonal copy of $H$ by the diagonal of $K$.

(B) Type B extension. There exists $(\alpha, \sigma) \in \operatorname{Aut}(H) \times S_t$ of order $p$ acting on $H \times \{1,\dots,t\}$ as above, such that the permutation $\pi(h,i) = (\alpha(h), \sigma(i))$ has the same fixed-point structure as the elements of $G$; then $L = \langle G, \pi \rangle$ satisfies the same uniform cycle-type property.

Answer 1

Since this has been up for a while and the bounty is about to run out with no answers, I'll write down some arguments suggesting the general answer probably involves difficult questions in extremal combinatorics.

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As noted, any $H \leq S_n$ with all non-identity elements of the same cycle type must be a $p$-group of exponent $p$ for some prime $p$. More precisely, the cycle type of all non-identity elements must be of the form $(p^k, 1^{n-pk})$ where $p, k$ depend only on $H$.

We'll take $p=2$ from now on, since it seems simpler. Groups of exponent $2$ are abelian (since $x^2 = y^2 = (xy)^2 = 1$ implies $xy = yx$), so indeed such an $H$ is an elementary abelian $2$-group, hence $H \cong (\mathbb{Z}/2)^m$ for some $m$. We won't actually use this observation, though I'll note it says you won't be able to find a counterexample to your distinguishing subgroups problem when $p=2$.

The following simple lemma describes when two elements of such an $H$ can possibly commute.

Lemma 1: Let $\sigma, \tau \in S_n$ be involutions with $\sigma \neq \tau$. Then $\sigma\tau = \tau\sigma$ if and only if for every $2$-cycle $(a, b)$ in $\sigma$, one of the following occurs:

1. $(a, b)$ is in $\tau$; or
2. $(a)(b)$ is in $\tau$; or
3. there is some unique $(c, d)$ disjoint from $(a, b)$ such that $(a, b)(c, d)$ is in $\sigma$ and $(a, c)(b, d)$ is in $\tau$.

Proof: Suppose $\sigma\tau=\tau\sigma$ and 1-2 do not occur. Then $\tau(a)=c$ for some $c \neq a, b$. Hence the portion of $\sigma = \tau\sigma\tau^{-1}$ involved in $a, b, c$ is of the form $(a,b)(c,?) = (\tau(a),\tau(b))(\tau(c),\tau(?)) = (c,\tau(b))(a,\tau(?))$. Hence they must be as described. Conversely 1-3 clearly imply commutativity. $\Box$

Lemma 2: If $\sigma, \tau \in S_n$ are commuting involutions of the same cycle type and $\sigma \neq \tau$, then $\sigma\tau$ has that same cycle type if and only if the number of cycles $(a, b)$ in both $\sigma$ and $\tau$ is the same as the number of cycles $(a, b)$ in exactly one of $\sigma$ or $\tau$.

Proof: From Lemma 1, we see quite directly when multiplying $\sigma\tau$ or $\tau\sigma$ that the number of cycles in case (3) is preserved. In case (1) they cancel and in case (2) we gain an extra cycle. Hence cases (1) and (2) must balance each other out. $\Box$

Proposition 1: Suppose $H \leq S_n$ and all non-identity elements have the same cycle type $(2^k, 1^{n-2k})$. Further suppose for any pair of involutions $\sigma, \tau \in H$ with $\sigma \neq \tau$, case 3 of Lemma 1 does not occur. Then $H$ is completely determined by a family of subsets $\mathcal{I}$ where for all $I, J \in \mathcal{I}$,

• $I \subset [N]$ where $N := \lfloor n/2\rfloor$,
• $|I| = k$, and
• $(I - J) \cup (J - I) \in \mathcal{I}$ if $I \neq J$.

Proof: By Lemma 1 and the assumption that case 3 does not occur, the collection of all $2$-cycles for all $\alpha \in H$ is some disjoint cycles $s_1, \ldots, s_m$. We may as well take $m=\lfloor n/2\rfloor=N$. Each non-identity $\alpha \in H$ is completely determined by the subset $I$ consisting of all $i$ for which $s_i$ is a cycle in $\alpha$. Clearly $|I|=k$, and by Lemma 2 we have $|I \cap J| = k/2$ for $\alpha \neq \beta$. Indeed, the set of cycles in $\alpha\beta$ corresponds to the symmetric difference $(I - J) \cup (J - I)$. Moreover the construction is completely reversible. $\Box$

Example 1: Consider the family of subsets $I_1 = \{1, 2, 3, 4\}, I_2 = \{3, 4, 5, 6\}, I_3 = \{1, 2, 5, 6\}$ of $[6]$. They correspond to involutions \begin{align*} &(1,2)(3,4)(5,6)(7,8) \\ &(5,6)(7,8)(9,10)(11,12) \\ &(1,2)(3,4)(9,10)(11,12), \end{align*} which generate a copy of the Klein 4-group in $S_{12}$.

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In Proposition 1, we're interested in a maximal family $\mathcal{I}$ of subsets of $[N]$ with the same fixed cardinality and which is closed under symmetric difference (of different subsets). This seems very restrictive, and I couldn't come up with much more than Example 1. But if we allow case 3 of Lemma 1 to occur, there is a lot more flexibility since the underlying ground set of possible $2$-cycles is no longer disjoint. If anybody has any further ideas, I'm happy to hear them.

Edit (as suggested in the comments): The condition in Proposition 1 is a variant of the setup in the Erdos--Ko--Rado theorem. You can use the $t$-intersecting generalization on that page there to give a bound on the cardinality of such $H$, though I imagine it's not a good one here since we want exactly half pairwise intersections. Nonetheless, it seems to me that the problem likely belongs in the realm of extremal combinatorics.