Let $\mathbf{V} = \mathbb{F}^n$ be a n-dimensional vector space over the field $\mathbb{F}$, and let $\sigma\in\mathbf{S}_n$ ($\mathbf{S}_n$ acts normally on $\mathbf{V}$). Suppose $\vec{v}\in \mathbf{V}$, and we have a list of vectors: $$B=[\vec{v},\sigma(\vec{v}),\sigma^2(\vec{v}),…,\sigma^k(\vec{v})]$$ Where $k$ is the period of $\sigma$ (the order of it’s orbit). Can we prove or disprove the existence of a proper subspace $U\subset\mathbf{V}$ generated by the list $B$ if $k=n-1$?
2026-03-31 05:02:32.1774933352
Least vector subspace generated by the orbit of a permutation.
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Your hypotheses seem to require that $\sigma \in S_n$ to be an $n$-cycle, and that the orbit $\{v, \sigma(v),\ldots, \sigma^{n-1}(v)\}$ of $v\in \mathbb F^n$ has size $n$?
Assuming this is correct, since $S_n$ acts via the permutation action on the standard basis $\{e_1,\ldots,e_n\}$ and we may assume that $\sigma$ is the permutation $(12\ldots n)$, if $v =\sum_{i=1}^n a_ie_i$, then the orbit of $v$ will have size $n$ provided that for each proper divisor $d$ of $n$ it is not the case that $a_i = a_j$ when $d$ divides $i-j$.
I think you are asking if it is necessarily the case that the orbit $B$ of $v$ spans $\mathbb F^n$? If that is correct,then the answer is ``no'': if we take $v= \sum_{i=1}^n (-s+2i)e_i$ where $s = n(n+1)$, then the coefficients of the $e_i$ form an increasing arithmetic progression, and so certainly the orbit of $v$ has size $n$. However it is also clear that the orbit lies in the proper subspace $H=\{\sum_{i=1}^n a_ie_i: \sum_{i=1}^n a_i =0\}$ of $\mathbb F^n$.
Of course $H$ is a subspace preserved by the action of all of $S_n$ (as is the line $\mathbb F(e_1+\ldots + e_n)$, which is a complement to $H$ unless $\text{char}(\mathbb F)$ divides $n$). One could, I suppose, also ask if there are vectors $v$ for which their orbit under $\sigma$ spans a subspace of $V$ which is not a subrepresentation of $S_n$, but only a subrepresentation of $C_n = \langle \sigma\rangle<S_n$ the cyclic group generated by $\sigma$. Here things depend on $\mathbb F$: if $\mathbb F$ contains an $n$-th root of unity, $\zeta$ say, then $v = \sum_{i=1}^n \zeta^{i-1}.e_i$ has $\sigma(v) = \zeta.v$, so that the orbit of $v$ has size $n$ but only spans a line in $V$, and the same will hold if we replace $\zeta$ by $\zeta^k$ if $k$ and $n$ are coprime.