Does there are exist simple proof for the following statement?
Let $\rho,V$ be an irreducible representation of group $G$ of dimention $n$. 
Assume that there are exist $g \in G$ such that $\rho(g)$ just flips two coordinates. (that is 
$\rho(g)e\_1=e\_2,\ \rho(g)e\_2=e\_1,\ \rho(g)e\_i=e\_i$) Then $|G|\geq 2^n$, where $n$ is a dimention of the representation. 



Irreducible representation flipping two elements
[I assume characteristic 0, and then I can extend the scalars to an algebraically closed field, which I can take to be $\mathbb{C}$ as $G$ is finite and $V$ is finite dimensional. It may appear that I then need the starting $V$ to be *absolutely* irreducible for the argument below, but actually since the eigenvector $e\_1 - e\_2$ is rationally defined, just irreducible is enough.]


$\rho(g)$ has a single $-1$ eigenvalue, and the rest are $1$, i.e. it is a (complex) symmmetry in the sense of Serre's book Complex Semisimple Lie Algebras. Let $v$ be a $-1$ eigenvector. Since $V$ is irreducible, the orbit of $v$ spans $V$. So the orbit forms a (complex) root system, where the symmetry corresponding to the root $\rho(h)(v)$ is $\rho(hgh^{-1})$ (note we've dropped the integrality hypothesis from Serre's definition of root system). Now complex root systems in this sense correspond to complexifications of real root systems. So the image under $\rho$ of the subgroup of $G$ generated by these symmetries is isomorphic to a (possibly reducible) finite real reflection group of rank $n$. Since reducible Coxeter groups are direct products of their parts, we need only check that irreducible finite Coxeter groups have order at least $2^r$, where $r$ is the rank, which is true by the classification, or by using the normal form for words in Coxeter groups to show that the $2^r$ words formed from omitting any subset of simple reflections from $s\_1 s\_2 s\_3 \dots s\_r$ give distinct elements.


