Linearly primitive implies cyclic-center

Verbal statement
If a group has a faithful irreducible representation over complex numbers, then the center of the group is cyclic.

Property-theoretic statement
The group property of having a faithful irreducible representation over complex numbers (viz the property of being a linearly primitive group) implies the group property of having a cyclic center (viz the property of being a cyclic-center group).

Symbolic statement
Let $$G$$ be a group with a faithful irreducible representation $$\rho$$ over complex numbers. Then, $$Z(G)$$ is a cyclic group.

Generalizations
It is in fact true that if a group has a faithful irreducible representation over any field of characteristic zero, then the center of the group is cyclic.

Proof
The proof follows from two facts.

Center goes to scalars
Schur's lemma states that for an irreducible representation $$\rho:G \to GL(V)$$, the only elements of $$GL(V)$$ that commute with everything in $$\rho(G)$$ are the scalar matrices.

Since the image of any element in $$Z(G)$$ must commute with everything in $$\rho(G)$$, we conclude that the image of any element in $$Z(G)$$ is scalar.

In particular, thus, we have a homomorphism from $$Z(G)$$ to the multiplicative group of complex numbers ($$\mathbb{C}^*$$).

In particular, if $$\rho$$ is faithful, $$Z(G)$$ is embedded as a subgroup of $$\mathbb{C}^*$$.

Any finite subgroup of multiplicative group of complex numbers is cyclic
We know that the finite subgroups of the multiplicative group of complex numbers are precisely the cyclic groups of $$d^{th}$$ roots of unity. These are always cyclic groups.

Combining these two facts, we obtain that the center is always cyclic.