Order of inner automorphism group bounds square of degree of irreducible representation

Statement with symbols
Let $$G$$ be a finite group and $$\rho: G \to GL(V)$$ an irreducible representation of $$G$$ over an algebraically closed field of characteristic zero. Let $$d$$ be the degree of $$\rho$$. Then:

$$d^2 \le [G:Z(G)]$$

Other facts about degrees

 * Degree of irreducible representation divides group order
 * Degree of irreducible representation divides index of center
 * Degree of irreducible representation divides index of abelian normal subgroup
 * Sum of squares of degrees of irreducible representations equals order of group
 * ‎Sum of squares of degrees of irreducible representations whose restriction to the center is a given character equals order of inner automorphism group

Breakdown for a non-algebraically closed field
Let $$G$$ be the cyclic group of order three and $$\R$$ be the field. Then, there are two irreducible representations of $$G$$ over $$\R$$: the trivial representation and the two-dimensional representation given via action by rotation by multiples of $$2\pi/3$$. The two-dimensional representation has degree $$2$$, and the square of its degree is $$2^2 = 4$$, which is greater than the order of the inner automorphism group, which is $$1$$.

The center cannot be replaced by an Abelian normal subgroup here
Notice that for an analogous result, namely, the degree of any irreducible representation divides the index of the center, we can strengthen the result to saying that the degree of any irreducible representation divides the index of any Abelian normal subgroup. However, the order-bound cannot be strengthened.

An example is the dihedral group of order eight, that has an irreducible representation of degree $$2$$. The square of this, which is $$4$$, is greater than the index of some of the Abelian normal subgroups.

Examples
In the examples below, we list, for each group, the degrees of its irreducible representations, the square of the maximum of these, the order of its center, the index of its center (which is also the order of the inner automorphism group). We omit the abelian groups, where both numbers are $$1$$.

Facts used

 * 1) uses::Irreducible representation over splitting field surjects to matrix ring
 * 2) uses::Schur's lemma

Proof
The proof below has the right ideas, but its presentation needs to be improved.

We use the crucial fact (called/coming from Schur's lemma) that the image of $$k(G)$$ under $$\rho$$ is the whole matrix ring $$Mat(V)$$, and that the image of the subalgebra $$k(Z(G))$$ is the one-dimensional subalgebra of scalars.

Clearly, the image of the elements in $$G$$ span $$Mat(V)$$ as a linear space. However, from the fact that $$Z(G)$$ maps to scalars, it follows that two elements in the same coset of $$Z(G)$$ are linearly related. Thus, by picking one representative from each coset, we can get a spanning set on $$Mat(V)$$ of size $$[G:Z(G)]$$.

Comparing this with the fact that the dimension of $$Mat(V)$$ is $$d^2$$, we obtain that:

$$d^2 \le [G:Z(G)]$$