Do sets of commuting permutations with no fixed points generate Abelian groups with no fixed points?

Question

Just the question

Let $S = \{\sigma_j\}_{j=0}^M$ be a set of permutations over $n$ elements that (i) have identity $\sigma_0 = I$, (ii) commute, and (iii) have no fixed points, i.e., $\sigma (i) \neq i$ for all $I \neq \sigma \in S$ and $i \in \{0, \ldots, n-1\}$.

Can we show that $\langle S \rangle$ is an Abelian group such that $I \neq \sigma \in \langle S \rangle$ has no fixed points? I'm pretty sure this is equivalent to saying $\langle S \rangle$ is also transitive (hence title and see also), but I can change the title if I am wrong.

Context

My background is in quantum information, and I am interested in translating and simplifying various arguments of the so-called "Permutation matrix representation" form (see Eq. 1 and Appendix A in arXiv:1908.03740v). A discussion beyond the link seems like an unnecessary digression, but I am happy to provide more context if desired. But as a meta point, my group theory is very primitive (hence my proof attempt below), but I don't mind answers that use sophisticated results---though I would request a link to a reference, if possible.

My attempt at a proof

We can show $\langle S \rangle$ is an Abelian group in two steps. First, let $A = \{\sigma_0^{k_0} \ldots \sigma_M^{k_M}: k_i$ non-negative integers$\}$ and $\sigma, \sigma' \in A$. Clearly, $\sigma \sigma' = \sigma' \sigma \in A$. Second, since $\sigma \sigma^{-1} = I$, then $\sigma \sigma^{-1} \sigma' = \sigma' \sigma \sigma^{-1}$ which implies $\sigma^{-1} \sigma' = \sigma' \sigma^{-1}$. Together, this shows we can write $\langle S \rangle = \{\sigma_0^{k_0} \ldots \sigma_M^{k_M}: k_i \in Z\}$, and hence, that $\langle S \rangle$ is an Abelian subgroup of $S_n$.

My problem is that my attempts to show $I \neq \sigma \in \langle S \rangle$ has no fixed points requires I already know the group is transitive and vice versa. In particular, suppose $\sigma$ has a fixed point. Then $\sigma \sigma' (i) = \sigma' (i)$ by assumption. If we can choose $\sigma'$ such that $\sigma'(i)$ hits all $k \in \{0, \ldots, n-1\}$, then we get that $\sigma = I$ which is a contradiction.

Since the above doesn't seem to work, is there an alternative approach, i.e., to show that $| \langle S \rangle | = n$ and that that + being Abelian is already enough? (see this post?) Or is my hope wrong from the start?

Relevant Stack Exchange posts

1. What does it mean to be a transitive permutation group?
2. Show that $\sigma(a)\ne a,\forall\sigma\in G-\{1\}$ and all $a\in A$.where $G$ is abelian, transitive subgroup of $S_A$
3. Show that any abelian transitive subgroup of $S_n$ has order $n$
4. Showing that a transitive abelian permutation group is necessarily regular
5. how many abelian transitive subgroups of $S_{n}$
6. Every Transitive Permutation Group Has a Fixed Point Free Element

Answer 1

It is difficult for permutation actions to have the property that no non-identity element has fixed points. This property is called being free, and for a group $G$ acting on a set $X$ it is equivalent to any of the following conditions:

1. Every orbit is isomorphic to $G$ acting on itself by left multiplication.
2. $X$ admits a cartesian product decomposition $G \times Y$ where $Y$ is some other set (which can be identified with the set of orbits) and $G$ acts on the left factor by left multiplication.
3. Every stabilizer subgroup $G_x = \{ g \in G : gx = x \}$ is trivial.

In particular, if $G$ and $X$ are both finite, a necessary condition for $X$ to be free is that $|G|$ divides $|X|$.

I don't know any conditions on a set of permutations guaranteeing that they generate a group that acts freely; basically the issue, as you've run into, is that it's hard to control how many fixed points a complicated product of your generators has. Abelianness doesn't really seem to make it easier either, as far as I can tell. If you'd like to make progress on this it would help to get a lot more specific about what specific sets of permutations you're interested in.

Edit: Here is another necessary condition. When a group acts freely every element $g$ of the group has to have a cycle decomposition in which every cycle has the same length, namely the order of $g$. So at a minimum the generators need to have this property.