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Let $ \pi: SU(n) \to PU(n) $ be the projection map.

Suppose that $ H_1 $ and $ H_2 $ are isomorphic subgroups of $PU(n)$. Must the subgroups $ \pi^{-1}[H_1] $ and $ \pi^{-1}[H_2] $ of $ SU(n) $ be isomorphic?

This is certainly true for $ n=2 $ since $ PU(2) \cong SO(3) $ and there is a one-to-one correspondence between the dihedral subgroups $ D_n $ of $ SO(3) $ and the binary dihedral subgroups $ BD_n $ of $ SU(2) $. Also the tetrahedral $ A_4 $, octahedral $ S_4 $ and icosahedral $ A_5 $ subgroups correspond to the binary tetrahedral $ 2.A_4 $, binary octahedral $ 2.S_4 $ and binary icosahedral $ 2.A_5 $ subgroups of $ SU(2) $ respectively. Even for cyclic groups in $ SO(3) $ there is a unique lift to a cyclic subgroup of $ SU(2) $ of twice the size.

Its not hard to extend the argument from finite $ H_i $ to all closed $ H_i $. The only infinite closed subgroups are $ SO(2) \cong U(1) $ in $ SO(3) $, which lifts uniquely to $ U(1) \cong SO(2) $ in $ SU(2) $, and then $ O(2) $ in $ SO(3) $ which lifts uniquely to a slightly weird subgroup of $ SU(2) $ which I call $ N $ in my answer https://math.stackexchange.com/a/4398724/724711 because it is the normalizer of the maximal torus of $ SU(2) $.

But does this hold in general? Already for $ PU(3) $ it is not quite as obvious to me that isomorphic subgroups of $ PU(3) $ must pull back to isomorphic subgroups of $ SU(3) $.

In general this just seems like a group extension problem $$ 1 \to \langle\zeta_n\rangle \to \pi^{-1}[H] \to H \to 1 $$ of $ H $ by a cyclic group of order $ n $, which should have many solutions, so perhaps pullbacks of isomorphic groups can be nonisomorphic?

Ian Gershon Teixeira
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  • Nice question, wish I knew more about $PU(n)$. If you knew there were more than one abstract group extension, then you couldn't rule out that only one of them fits in $SU(n)$, although in the unlikely event that your sequence is forced to split (which someone who knows more about $PU(n)$ can presumably rule out) then in fact the desired statement is true. – hunter Sep 09 '24 at 17:39
  • Based on your argument for the $n=2$ case, are you only considering the $H_i$ finite? – Steve D Sep 09 '24 at 18:22
  • @SteveD ok I extended my argument for the $ n=2 $ case to all closed subgroups $ H_i $ – Ian Gershon Teixeira Sep 09 '24 at 21:05

2 Answers2

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They don't have to be, but the counterexample I have is over finite fields:

$PSU(4,3)$ has two conjugacy classes of cyclic subgroups of order $4$. Their pre-images in $SU(4,3)$ have structure $C_4\times C_4$, respectively $C_8\times C_2$.

ahulpke
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  • +1 Ok this counterexample is very interesting. Is there a way to turn this into an example for $ SU(4) $? Can we be sure that counterexamples exist for the complex case? – Ian Gershon Teixeira Sep 10 '24 at 17:59
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Ok I did some more thinking inspired by the answer from ahulpke and also this answer.

One counterexample is the following. Consider projective irreps of $ A_5 $. These correspond to the regular irreps of the Schur covering group of $ A_5 $, which is $ SL(2,5)\cong 2.A_5 $. There are two irreps of $ SL(2,5) $ of degree $ 4 $. The first irrep is not faithful and its image is the $ A_5 $ subgroup of $ SO(4) \subset SU(4) $ and the the second irrep is faithful and its image is the $ SL(2,5) $ subgroup of $ Sp(2) \subset SU(4) $. Passing to $ PU(4) $ we quotient out the center and we have that the subgroups $ H_1=\pi(A_5) $ and $ H_2=\pi(SL(2,5)) $ of $ PU(4) $ are both isomorphic to $ A_5 $.

But pulling back we have that $ \pi^{-1}[H_1] \cong A_5 \times 4 $ (this is SmallGroup(240,92) in GAP) is a direct product of $ A_5 $ with a cyclic group of order $ 4 $ while $ \pi^{-1}[H_2] \cong SL(2,5) \circ_2 4 $ (this is SmallGroup(240,93) in GAP) is the central product of $ SL(2,5) $ with cyclic group $ 4 $ over the shared central cyclic $ 2 $ group.

Thus the subgroups $ H_1,H_2 $ of $ PU(4) $ are isomorphic but the the subgroups $ \pi^{-1}[H_1], \pi^{-1}[H_2] $ of $ SU(4) $ are not isomorphic.

Ian Gershon Teixeira
  • 10,176