Irreducible representation and dimension of image space Let $G$ be a finite group, $F$ a field of characteristic $0$ or of relatively prime to $|G|$.
Let $\rho:G\rightarrow {\rm GL}_n(F)$ be an irreducible representation of $G$.
Extend $\rho$ to algebra homomorphism $\rho_1:F[G]\rightarrow M_n(F)$. 
(*) If $F$ is algebraically closed then $\rho_1$ is surjective..
(**) If $F$ is not algebraically closed, then $\rho_1(F[G])$ is not necessarily $M_n(F)$ but it is certainly a subspace of $M_n(F)$. 
Example: Let $\rho:\langle x:x^4\rangle\rightarrow {\rm GL}_2(\mathbb{Q})$ be given by $$\rho(x)=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}.$$
We can see that $\rho(1)=I, \rho(x^2)=-I$ and $\rho(x^3)=-\rho(x)$. Thus, among the images of $\rho$,only $\rho(1)$ and $\rho(x)$ are independent in $M_2(\mathbb{Q})$, and so spans a subspace of dimension $2$ in $M_2(\mathbb{Q})$.
This forced me to think about: What this $2$ has to do with $G=\langle x:x^4=1\rangle$? 
To state question:
If $\rho_1(F[G])$ is proper subspace of $M_n(F)$, how is its dimension is related to group $G$? Is this dimension of image of $F[G]$ well studied for irreducible representations?
 A: In this case we have the dimension of the second $4$th cyclotomic polynomial $\Phi_4(X)=X^2+1$. For cyclic groups of order $n$, one can construct a representation with dimension is the degree of the $k$th cyclotomic polynomial for any $k|n$. This follows from factoring 
$$X^n-1=\prod_{k|n}\Phi_k(X).$$
In the case at hand, observe that
$$\mathbb{Q}G\cong\mathbb{Q}[X]/(X^4-1)\cong\mathbb{Q}[X]/(X-1)\oplus\mathbb{Q}[X]/(X+1)\oplus\mathbb{Q}[X]/(X^2+1),$$
where the last isomorphism is the Chinese remainder theorem. The representation in question is obtained by letting $x\in G$ act on the last factor by multipication by $X$. It is two dimensional because $\deg(X^2+1)=2$. (The other two representations of this group are one dimensional with $x$ acting $\pm1$ and come from the other two factors.)
Observe that when $G=\langle x\mid x^3=1\rangle$, then 
$$\mathbb{Q}G\cong\mathbb{Q}[X]/(X^3-1)\cong\mathbb{Q}[X]/(X-1)\oplus\mathbb{Q}[X]/(X^2+X+1).$$
Let $G$ act on the last factor $V=\mathbb{Q}[X]/(X^2+X+1)$ so that the generator acts by left multiplication by $X$. A basis for $V$ is $\{1,X\}$ and $X^2=-X-1$, so the matrix for left multiplication by $X$ is
$$\rho(x)= \begin{bmatrix}0&-1\\1&-1\end{bmatrix}.$$
Sure enough, $$\rho(x^2)=\begin{bmatrix}-1&1\\-1&0\end{bmatrix}=-\rho(x)-\rho(1),$$
and $\rho(x^3)=I$. Hence, we have obtained a two dimensional representation of $G$, the dimension corresponding to the degree of $\Phi_3(X)=X^2+X+1$. 
