Let $G$ be the group with center $Z_G$. By Schur's lemma $Z_G$ acts by scalar matrices. Thus we have an irreducible projective representation of the quotient $G/Z_G$ which is a finite group. Now, your question follows from the following fact:

Proposition Let $\Gamma$ be a finite group and let $V$ be a finite-dimensional projective representation of $\Gamma$. Then $\dim(V)$ divides the order $|\Gamma|$.

Proof It is known (and I think goes back to Schur) that any projective representation $V$ of $\Gamma$ lifts to a representation of some finite central cover $\widetilde\Gamma$ of $\Gamma$. Now $\dim(V)$ divides the order of $|\widetilde\Gamma/Z_{\widetilde\Gamma}|$ as proved in e.g. Section 6.4 of Serre's book. But $\Gamma$ maps surjectively onto $\widetilde\Gamma/Z_{\widetilde\Gamma}$ so $\dim(V)$ also divides $|\Gamma|$.

Let $G$ be the group with center $Z_G$. By Schur's lemma $Z_G$ acts by scalar matrices. Thus we have an irreducible projective representation of the quotient $G/Z_G$ which is a finite group. Now, your question follows from the following fact:

Proposition Let $\Gamma$ be a finite group and let $V$ be a irreducible finite-dimensional projective representation of $\Gamma$. Then $\dim(V)$ divides the order $|\Gamma|$.

Proof It is known (and I think goes back to Schur) that any projective representation $V$ of $\Gamma$ lifts to a representation of some finite central cover $\widetilde\Gamma$ of $\Gamma$. Now $\dim(V)$ divides the order of $|\widetilde\Gamma/Z_{\widetilde\Gamma}|$ as proved in e.g. Section 6.4 of Serre's book. But $\Gamma$ maps surjectively onto $\widetilde\Gamma/Z_{\widetilde\Gamma}$ so $\dim(V)$ also divides $|\Gamma|$.