Dimension of a cyclic submodule of a finite group representation

Let $\rho: G \to GL(V)$ be a finite (complex) group representation. What is the maximum dimension of $span\{\rho_gv | g \in G\}$ over all $v \in V$? This quantity is not necessarily the degree of $V$: for example, if $V$ is the direct sum of two isomorphic 1-dimensional subrepresentations, then the maximum dimension is 1.

I think this can be rephrased as: given a $\mathbb{C}G$-module, what is the largest (vector space) dimension over $\mathbb{C}$ of a cyclic submodule?

What I have so far: by the canonical decomposition, $V \cong W_1 \oplus W_2 \oplus ... \oplus W_r$ where each $W_i$ is isomorphic to $n_i$ copies of an irreducible submodule $U_i$ with dimension $d_i$. For $v \in V$ define $\Phi = \sum_{g \in G}{\rho_gvv^*\rho_g^*}$ (* denotes adjoint). Then the dimension quantity in question is equal to $rank(\Phi)$. $\Phi$ is a $\mathbb{C}G$-module homomorphism of $V$ into itself. Therefore $\Phi$ maps each $W_i$ to itself. Hence if $v = w_1 + w_2 + ... + w_r$ is the direct sum decomposition of $v$, then the dimension of $span\{\rho_g{v} | g \in G\} = \sum_{i = 1}^r{dim (span\{\rho_gw_i | g \in G\})}$, thus reducing the problem to the situation where the representation in question is isomorphic to copies of an irreducible module.

If this irreducible module has degree 1, then the maximum dimension is 1. But for degree > 1, I'm not sure. I haven't been formally taught representation theory, so I'd appreciate references as well as a clue to this problem.

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Isn't it true that the span of $\{\rho_g v : g \in G, v \in V\}$ is equal to the span of $\{\rho_g v : g \in G, v \in B\}$ where $B$ is any basis of $V$? If $G$ is finite, then you can compute the dimension of this latter span by checking for linear independence of the vectors in $\{\rho_g v : g \in G, v \in B\}$. –  echoone Jul 4 '12 at 19:36
I think you're right, except I'm looking for the span of $\{\rho_gv : g \in G\}$ for a particular v. If it is the span over all $v \in V$ or over a basis of $V$, then that space is the full space V. And while I can check linear independence for a particular $v$ and $G$, I'm trying to express the dimension generally in terms of the irreducible components (which might require finding a particular $v$ that maximizes the dimension). –  Daniel Copeland Jul 4 '12 at 19:51
Your question is about the dimension of the cyclic $\mathbb{C}G$-module generated by $v$. Such a module is obviously isomorphic to a quotient of the left regular representation $M=\mathbb{C}G$ (= the free cyclic module generated by $1$), as there is a unique homomorphism $f:M\rightarrow \langle v\rangle$ determined by $f(1)=v$. Conversely any such quotient is a cyclic module as a homomorphic image of $M$ is generated by the image of $1$.
Let us denote by $m(V,U_i)$ the multiplicity of an irreducible representation $U_i$ as a composition factor of $V$. For each $U_i$ we know that $m(M,U_i)=\dim U_i$. Therefore the maximum multiplicity of $U_i$ as a composition factor of a cyclic module is also equal to $\dim U_i$. Conversely (using Maschke's theorem), if a module $V$ satisfies the inequality $m(V,U_i)\le \dim U_i$ for all irreducible representations $U_i$, it can be written as a quotient of $M$, i.e. is a cyclic module.
Therefore the answer is that the maximum dimension of a cyclic submodule of a given module $V$ is $$\sum_i\min\{\dim U_i, m(V,U_i)\}\cdot\dim U_i.$$ The summation ranges over the irreducible representations of $G$.