About an irreducible representation over an algebraically closed field I want to prove the following statement that is an of the book "A course in the theory of groups" by D. Robinson:
Let $n$ be the degree of an irreducible representation of a finite group $G$ over an algebraically closed field. Prove that $n^2\le \vert G:Z(G)\vert$.
The hint of this exercise is: "Use the following theorem"

I can prove this statement over the field of complex numbers, but over any algebraically closed fields, I don't have any useful ideas.
Thanks in advance.
 A: Let $k$ be an algebraically closed field and $A$ a finite-dimensional $k$-algebra. The Artin-Wedderburn Theorem tells us that $\dim_k(S)^2 \leq \dim_k(A)$ for all simple $A$-modules $S$. 
Now, if $S$ is a simple $kG$-module, then by Schur's Lemma there exists a $\theta : Z(G) \to k^*$ such that $z x = \theta(z) x$ for all $z \in Z(G)$ and $x \in S$. Hence, $S$ becomes a simple $kG/I$ -module where $I = \langle \theta(z)1 - z \;|\; z \in Z(G) \rangle$. Note that a basis of $kG/I$ is given by representatives for the cosets of $G/Z(G)$, hence $\dim_k(kG/I) = |G:Z(G)|$ and putting things together we obtain $\dim_k(S)^2 \leq \dim_k(kG/I) = |G:Z(G)|$.
EDIT:
Using the theorem of Burnside, we can prove the statement in the following way.
Let $S$ be a simple $kG$-module and $\rho : kG \to \text{End}_k(S), g \mapsto (x \mapsto x s) $ the corresponding representation. Then the theorem tells us that this map is surjective.
Now, similarly to the first proof we have $I \subseteq \ker(\rho)$ and so we get an induced homomorphism $kG/I \to \text{End}_k(S)$ which is also surjective. This gives $$|G:Z(G)| = \dim_k(kG/I) \geq \dim_k \text{End}_k(S) = \dim_k(S)^2.$$
