Dimension of an irreducible representation and index of the center G is a finite group of order n, Z(G) is its center of order 'k'. I want to show that if V is an irreducible complex representation of G, then dim(V) $\leq$ $\sqrt{\frac{n}{k}}$. 
I noted that Z(G) is an abelian subgroup of G and with a little effort was able to arrive that dim(V) $\leq$ [G : Z(G)] = $\frac{n}{k}$. However, I don't see a way to get to the actual result. Any hints would be appreciated.  
 A: Let $\rho : G \to GL(V)$ be a complex irreducible representation. Extending this, we obtain an algebra morphism $\rho : \mathbb{C}G \to \text{End}_\mathbb{C}(V)$. Now you can proceed by showing the following:


*

*The algebra morphism $\rho$ is surjective (this is a consequence of the Artin-Wedderburn theorem, since $\mathbb{C}G$ is semisimple). In particular, 
$$\dim(\mathbb{C}G/\ker(\rho)) \geq \dim(\text{End}_\mathbb{C}(V)) = \dim(V)^2$$

*The elements of $Z(G)$ act on $V$ as scalars (by Schur's Lemma).
In particular $\{\lambda(z) - z \:|\: z \in Z(G)\}$ is contained 
in $\ker(\rho)$ for some $\lambda : Z(G) \to \mathbb{C}^*$.

*The quotient $\mathbb{C}G/\ker(\rho)$ is generated by the cosets of $Z(G)$ in $G$ and so $$\dim(\mathbb{C}G/\ker(\rho)) \leq n/k.$$
Putting the inequalities in 1. and 3. together yields the result.
A: Here is another method using the tensor power trick, based on an argument of Tate.
By Schur's lemma, each element $z \in Z(G)$ acts on $V$ by a scalar $\lambda_z \in \mathbb{C}$. Assume for the moment that $Z(G)$ acts trivially on $V$, i.e. $\lambda_z = 1$ for all $z \in Z(G)$. Then $V$ is an an irrep of the quotient group $G/Z(G)$. Since the size of a finite group equals the sum of squares of the dimensions of its irreps, we conclude that $\dim(V)^2 \le |G|/|Z(G)|$, as desired.
While it may not be the case that $Z(G)$ acts trivially on $V$, we can modify the above strategy by passing to the tensor power $V^{\otimes n}$, which is an irrep of $G^n$. An element $(z_1, \ldots, z_n) \in Z(G)^n$ acts on $V^{\otimes n}$ by multiplication by $\lambda_{z_1} \cdots \lambda_{z_n}$. Therefore, the normal subgroup $$ H_n := \{(z_1, \ldots, z_n) \in Z(G)^n \mid z_1 \cdots z_n = 1\} $$ acts trivially on $V^{\otimes n}$. Thus, $V^{\otimes n}$ is an irrep of the quotient group $G^n/H_n$. We conclude that $$\dim(V)^{2n} = \dim(V^{\otimes n})^2 \le |G^n/H_n| = \frac{|G|^n}{|Z(G)|^{n - 1}}.$$ Taking $n$th roots and sending $n \to \infty$, we conclude that $\dim(V)^2 \le |G|/|Z(G)|$.
