Consider the trace map $tr: M_{n \times n} ( \mathbb{F}) \to \mathbb{F}$. It is well-known that this satisfies the property:
$$tr(AB) = tr(BA)$$
This is equivalent to $tr(A_1 \cdots A_n) = tr(A_{\sigma(1)} \cdots A_{\sigma(n)})$ if and only if $\sigma = (1 \cdots n)^k$ for some $k$ (it is easy to mistakenly believe that it implies $\sigma$ can be arbitrary).
If we abstract the underlying algebra here, there is some sense in which the abelian structure of $C_2 \le S_2$ induces the subgroups $C_n \le S_n$ (specifically, $C_n = \langle (1...n) \rangle$, which I omit from now on) for all $n$. To expand on this, consider the tuple:
$$(1, ..., n)$$
We can consider partitioning this into two tuples in an order-preserving way:
$$(1, ..., m)(m+1,...,n)$$
We consider $C_2$ acting by rearranging these tuples to get:
$$(m+1, ..., n)(1, ..., m)$$
In this way, I am stating that $C_2$ may be considered to induce the permutation $$\tau = (m+1, ..., n, 1, ..., m) = (1, ..., n) \in S_n$$
In a similar way, it is possible to consider a subgroup $H \le S_m$ inducing subgroups $H_n \le S_n$ for each $n \ge m$ -- the span of all permutations generated by breaking down $(1, ..., n)$ into $m$ sub-tuples and considering the permutation corresponding to the action of $\tau \in H$ on these sub-tuples.
We will also describe a sort of dual operation to this. Let $H \le S_m$. Then, $H$ induces subgroups $H_n \le S_n$ for all $n \le m$ as follows. Given some $\sigma \in H$, let $i_1, ..., i_n \in \{1, ..., m\}$ be pairwise distinct. Then, I state that $\sigma$ induces the permutation:
$$ 1 \mapsto \min_j {\sigma(i_j)}$$ $$\vdots$$ $$n \mapsto \max_j {\sigma(i_j)}$$
Let $H_n \le S_n$ be the span of all elements $\tau$ induced in this way.
Definition: The sequence $(G_k \le S_k)$ is consistent if, for every $k$, the subgroup of $S_n$ induced by $G_k$ (by the first operation if $n \ge k$, or its dual if $n \le k$) is contained in $G_n$.
Question: Is it possible to have a consistent sequence with $C_k < G_k < S_k$ (i.e. proper subgroup inclusions) for some $k$?
- $ \exists k : G_k \neq \{e\} \implies C_k \le G_k \forall k$
- $ \exists k \ge 3 : G_k = S_k \iff \forall k, G_k = S_k$
- $ \exists k \ge 4 : G_k = A_k \implies \forall k, G_k = S_k$
I will not write the proofs for these. None are difficult.
I have found three examples of consistent sequences:
- $G_k = \{e\} \forall k$
- $G_k = C_k \forall k$
- $G_k = S_k \forall k$
By the first result I stated, there are none "in between" (1) and (2). It would be of great interest to me to find a sequence between (2) and (3).
More generally, it would be helpful to know about any results on the structure of groups $G$ where $\langle (1...n) \rangle \le G \le S_n$.
Proposition
Every consistent sequence is of the form $G_k \equiv \{e\}, C_k$ or $S_k$.
Proof (sketch)
We prove that $(\exists n \ge 3: C_n \le G_n, \; C_n \neq G_n) \implies (G_3 = S_3) \implies (G_k = S_k \; \forall k)$.
Firstly, observe that if $e \neq \sigma \in G_n$, then $\sigma$ cannot have any fixed points. Otherwise, let $1 \in \{1,2,3\}$ be fixed under $\sigma$. Since $\sigma$ is not the identity, there must exist an induced permutation which interchanges $2,3$. (This is easy to see if one considers the disjoint cycle decomposition of $\sigma$.) In particular, if $\sigma$ were to have a fixed point, then we would have $(23) \in G_3 \implies G_3 = S_3$.
Hence, every $e \neq \sigma \in G_n$ must not have any fixed points i.e. every non-identity element of $G_n$ must be a derangement. Recall that $(1...n) \in G_n$. Note that for any $\sigma \in G_n$, the permutation
$$(1...n)^{1-\sigma(1)} \sigma \in G_n$$
has $1$ as a fixed point. Thus, $(1...n)^{1-\sigma(1)} \sigma = e$.
$$\implies \sigma = (1...n)^{\sigma(1)-1}$$
$$\implies \sigma \in \langle (1...n) \rangle$$
Therefore, the only group of derangements $C_n \le G_n \le S_n$ is $C_n$.
We conclude that $(\exists n \ge 3: C_n \le G_n, \; C_n \neq G_n) \implies G_k = S_k \; \forall k$. $\square$
I may as well add why this was interesting to me.
Lemma
$N \le G$ is normal if and only if $xy \in N \implies yx \in N$.
One can also show that:
Lemma $$(xy \in N \implies yx \in N) \iff (\forall n : x_1 \cdots x_n \in N \implies x_{\sigma(1)} \cdots x_{\sigma(n)} \in N \text{ for } \sigma \in \langle (1...n) \rangle)$$
This gives a natural reason to define $(G_n)$ and the notion of consistency as above. The other two possible consistent sequences correspond to the cases given below:
Proposition
Corollary
Let $G$ be a non-abelian group. Then, for every $n$, $\sigma \in S_n \setminus C_n$, $\exists x_1, ..., x_n \in G$ such that:
$$x_1 \cdots x_n = e \; \; \text{ but } \; \; x_{\sigma(1)} \cdots x_{\sigma(n)} \neq e$$
Proof
$G / \{e\}$ is not abelian. $\square$