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Let $G$ be a group with a homomorphism $\varphi$ into $GL(V)$ for some finite dimensional vector space $V$ of dimension $n$. Let $d \leq n$ be some positive integer.

When does there exist a list of $d$ elements $a_i \in G$ such that, if $$\sigma^i = \sigma_i \circ \dots \circ \sigma_1$$ where $\sigma_i = \varphi (a_i)$, then $\exists v\in V $ such that $\{\sigma^i(v)| 0 \leq i \leq d\}$ is a set of linearly independent vectors?

My guess is that this is related somehow to semisimple representations. Specifically if we impose that $\varphi$ be semisimple, such that we can write a semisimple representation as a direct sum of at least $d$ subrepresentations, then it seems that we can get the desired property, but I am not sure.

Are there any theorems/papers in this area which might be helpful at looking at this?

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Suppose that $V$ is a cyclic $n$-dimensional representation of $G$, namely $V=span(G\cdot v)$ for some $v\in V$. Since the set $\{gv\mid g\in G\}$ is spanning, you can find $\{g_1,...,g_n\}$ such that $g_1v,...,g_nv$ is a basis, and in particular it is linearly independent. On the other hand, if you can find such $g_i$, then the representation must be cyclic. With your notations, you have that $a_1=g_1$ and $a_i = g_i (g_{i-1})^{-1}$ for $i>1$.

If you want the same result for some $d<n$, then the corresponding condition is that $span\{Gv\}$ is $m$-dimensional for some $v\in V$ and $m\geq d$. If the representation is simple, then $span\{Gv\}=V$ for any nonzero vector, but cyclicity is a weaker condition. For example, the regular representation of a finite group $G$ on the group algebra $V=\mathbb{C}G$ is not simple (unless $G$ is trivial) and you still have $V=span\{G\cdot 1\}$. More over, a representation $V$ is a cyclic representation if and only if it is a quotient of the regular (free) representation.

Also, semisimplicity is not a good condition. In particular, if $G$ is finite, then any finite dimensional representation is semisimple (assuming that you're not working is positive characteristic). If you take the trivial action of $G$ on an $n$-dimensional space $V$ (which is a direct sum of $n$ copies of the trivial representation), then $\{g_i(v)\}=\{v\}$ for any choice of $g_i$.

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