Cyclic vector for a group representation

A vector $v$ is cyclic for a group representation $\rho :G\to GL(V)$ if the linear span of $\{\rho (g)v:g\in G\}$ is all of $V$. Equivalently, $v$ generates $V$ as a module over the group algebra. For a finite group over a field in which $|G|$ is invertible, the averaging projection operator $P=|G{|}^{-1}{\sum }_g\rho (g)$ has image the fixed-point subspace of a group action. Since $P\rho (g)v=Pv$, cyclicity implies $V^G=kPv$, so this space has dimension at most one.

The invertibility hypothesis matters. In characteristic $p$, let $G=C_p\times C_p$ and let $V=k[x,y]/(x,y{)}^2$, with the two generators acting by multiplication by $1+x$ and $1+y$. The vector $1$ is cyclic, but both $x$ and $y$ are fixed. Thus a cyclic group representation in modular representation theory can have a two-dimensional fixed-point subspace of a group action.

In a unitary representation of a finite group, if a cyclic vector $v$ satisfies $\rho (g)v=av$ for some $a\ne 1$, then there are no invariant vectors. Indeed the averaging orthogonal projection $P=|G{|}^{-1}{\sum }_h\rho (h)$ satisfies $P\rho (g)=P$, so $Pv=aPv=0$. All translates of $v$ also project to zero, and they span the representation.