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Suppose we have a group representation $$\rho: G \to GL(V)$$ where $V$ is a finite-dimensional complex vector space and $|G| < \infty$. I have been confusing myself about the definition of multiplicity and the trivial representation.

Let $$V=\bigoplus_{\lambda}n_{\lambda}V^{\lambda}$$ be a decomposition of $V$ into irreducible representations, where $n_{\lambda}$ is the multiplicity of $V^{\lambda}$. Let $V_0=n_{triv}V^{triv}$ be the multiplicity of the trivial representation, and let $v \in V$ be a vector such that for all $g \in G$ we have $$\rho(g)(v)=v.$$ Does it necessarily follow that $v \in V_0$? Is it true that $$V_0=\{v \in V : \forall g \in G, \; \rho(g)(v)=v\}?$$

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Okay, I think I cleared up the confusion. Let $B_{\lambda}$ be a basis for each multiplicity. Then the union of all the $B_{\lambda}$'s is a basis $B$. Suppose there exists a $v \in V$ such that $v \not\in V_0$. Then $$v= \sum_i c_ib_i$$ where $c_i \neq 0$ for some $b_i \in B_{\lambda}$ where $\lambda \neq triv$. But this means the set $B_{triv} \cup \{v\}$ is linearly independent and so $\mathbb{C} \cdot v$ is an invariant subspace of $V$, implying $n_{triv} > n_{triv}$, a contradiction. Does that work?

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