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$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ G $ be a finite subgroup of $ \GL(n, \mathbb{C}) $ (and to nail things down a little more also suppose that $ G $ is irreducible in this natural representation). Let $ [G] $ be the corresponding subgroup of $ \PGL(n,\mathbb{C}) $.

For every degree $ n $ projective irrep $ \rho $ of $ [G] $ does there exist a subgroup $ \tilde{G} $ of $ \GL(n, \mathbb{C}) $ such that $ \tilde{G} $ is a Schur cover of $ [G] $ and moreover $ \rho $ lifts to a linear irrep $ \tilde{\rho} $ of $ \tilde{G} $?

This can be viewed as a question in two parts. (1) If $ [G] $ is an irreducible finite subgroup of $ \PGL(n,\mathbb{C}) $ does there always exist a Schur cover $ \tilde{G} $ of $ [G] $ that is a subgroup of $ \GL(n,\mathbb{C}) $? (2) Can any degree $ n $ projective irrep of $ [G] $ be lifted to a linear irrep of some such Schur cover of $ [G] $?

This seems to be true for $ n=2 $, where the only irreducible finite subgroups of $ \PGL(2,\mathbb{C}) $ are the dihedral groups $ Dih(n) $ of order $ 2n $ (with Schur covers $ Dih(2n) $ and $ Dic(n) $, both subgroups of $ \GL(2,\mathbb{C}) $ ), and then $ A_4\cong \PSL(2,3) $ (with unique Schur cover the binary tetrahedral subgroup of $ \GL(2,\mathbb{C}) $, isomorphic to $ \SL(2,3) $ ), $ S_4 \cong \PGL(2,3) $ (with two Schur covers: $ \GL(2,3) $ and also the binary octahedral subgroup of $ GL(2,\mathbb{C}) $), $ A_5\cong \PSL(2,5) $ (with unique Schur cover the binary icosahedral subgroup of $ \GL(2,\mathbb{C}) $, isomorphic to $ \SL(2,5) $).

Side note showing that $ \GL(2,3) $ really is (isomorphic to) a subgroup of $ \GL(2,\mathbb{C}) $:

let $ S= \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} $, $ H= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $. Then $$ \langle\overline{\zeta_8} S, iH\rangle $$ is a the binary octahedral subgroup of $ \SU(2) $. While $$ \langle\zeta_8 S, H\rangle $$ is the $ \GL(2,3) $ subgroup of $ \SU^{\pm 1}(2) $, the groups of unitary matrices with determinant $ \pm 1 $. (shout out to this excellent answer https://math.stackexchange.com/a/3670522/724711 which sort of inspired this choice of generators)

Ian Gershon Teixeira
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    The answer to (1) is no if the Schur Multiplier of $[G]$ is not cyclic. I think the answer to(2) is yes. – Derek Holt Nov 24 '24 at 21:33
  • @DerekHolt what do you mean by "the answer to (2) is yes"? Like does that mean that the Schur multiplier being cyclic is the only obstruction and if $ [G] \subset PGL(n,\mathbb{C}) $ has cyclic Schur multiplier then there always exist Schur cover $ \tilde{G} $ in $ GL(n,\mathbb{C}) $ lifting any degree n projective rep? And as for your point about needing Schur multiplier cyclic that is a really good point I should have realized that would be important, I'll look in GAP for a specific counterexample with like maybe $ PSL(3,4) $? What are some small groups with non-cyclic Schur multiplier? – Ian Gershon Teixeira Nov 24 '24 at 22:05
  • Yes $G={\rm SL}(3,4)$, $[G]={\rm PSL}(3,4)$ with $n=15$ gives a counterexample. There are probably smaller ones. – Derek Holt Nov 24 '24 at 22:19
  • $GL = G\cdot L$ so your $\LaTeX$ is unreadable. – Daniel Donnelly Nov 24 '24 at 22:30
  • @DerekHolt Ah ok so if I understand you correctly then you are saying that (2) is correct in the sense that every degree n projective irrep of [G] arise from a linear irrep of some stem extension $ \tilde{G} $ of [G] with $ \tilde{G} $ a subgroup of $ PGL(n,\mathbb{C}) $. However these subgroups of $ PGL(n,\mathbb{C}) $ that are stem extensions of [G] may not be Schur covering groups (i.e. they may not be stem extensions with the maximal size among all finite groups). – Ian Gershon Teixeira Nov 24 '24 at 22:31
  • And then the counter example you are suggesting for (1) is that $ PSL(3,4) $ is a subgroup of $ PGL(15,\mathbb{C}) $ but the Schur cover of $ PSL(3,4) $ cannot be a subgroup of $ GL(15,\mathbb{C}) $ since the Schur multiplier of $ PSL(3,4) $ is not cyclic – Ian Gershon Teixeira Nov 24 '24 at 22:34

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