$\DeclareMathOperator\PSU{PSU}$Let $ \PSU_n $ be the projective unitary group. Let $ A_m $ be the alternating group on $ m $ letters.

$ A_5 $ is a maximal closed subgroup of $ PSU_2 \cong SO_3(\mathbb{R}) $.

The references in The finite subgroups of SU(n) show that $ A_6 $ is a maximal closed subgroup of $ \PSU_3 $.

The reference https://arxiv.org/abs/hep-th/9905212 from the same MO question shows that $ A_7 $ is a maximal closed subgroup of $ \PSU_4 $. That leads me to ask: Is $ A_{n+3} $ always a maximal closed subgroup of $ \PSU_n $?

To see maximality in these first three cases it is enough to realize that these subgroups are all 2-designs (in fact all 3-designs) and are maximal finite.

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    $\begingroup$ According to R. Brauer (Über endliche lineare Gruppen von Primzahlgrad, Math. Ann. 169, 73-96 (1967), $A_8$ is not a subgroup of $PSU(5)$. $\endgroup$
    – abx
    Jan 20, 2022 at 5:07

1 Answer 1


The comment of @abx is part of a general picture. For $n >7$ the alternating group $A_{n}$ has no non-trivial complex irreducible character of degree less than $n-1$, so that for $n > 7$, $A_{n}$ is not isomorphic to any subgroup of ${\rm GL}(n-3, \mathbb{C}).$

Furthermore, if $n > 7$ is also even, then the double cover of $A_{n}$ (which, by a theorem of I. Schur is a maximal perfect central extension of $A_{n}$) has no faithful complex irreducible character of odd degree, so certainly not of degree $n-3$ ( as an involution in its centre would be represented by a matrix of determinant $-1$). Hence the double cover of $A_{n}$ has no non-trivial irreducible complex representation of degree $n-3$, faithful or not. Thus $A_{n}$ does not embed as an irreducible subgroup of ${\rm PSU}(n-3, \mathbb{C})$ (and it is clear that a maximal closed subgroup has to be irreducible).

(A few points for clarity: for finite groups, all finite dimensional complex representations are equivalent to unitary ones. Also, there is no real distinction between projective representations (in Schur's sense) and genuine representations of covering groups. The fact that the minimal degree of an non-triival irreducible complex representation of $A_{n}$ is $n-1$ ( for $n > 7$) can be found in almost any text on representation theory of $S_{n}$ ( eg that of G.D. James)).

Later edit: I am not sure of the smallest degree of a faithful projective (in Schur's sense) complex representation of $A_{n}$, though I believe this must be well-known to symmetric group experts. However, results such as Zsygmondy's theorem provide a lower bound. If $p$ is a prime less than or equal to $m-2$, then $A_{m}$ contains a $p$-cycle which is conjugate to all its non-identity powers, from which it follows that $A_{m}$ can have no non-trivial complex projective representation of degree less than $p-1.$ Hence if $n$ is an even integer such that there is a prime $p$ with $\frac{n}{2} < p < n-2$, we see that $A_{n}$ has no complex irreducible projective representation of degree less than $\frac{n}{2}$ (and we may find similar bounds for odd $n$).

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    $\begingroup$ I have an idea for a reformulation of the question but this answer is so lovely I think it is best practices for me to just accept this answer and make a new post. $\endgroup$ Jan 20, 2022 at 16:18
  • $\begingroup$ Here is the new post mathoverflow.net/questions/414315/finite-simple-groups-and-su-n $\endgroup$ Jan 20, 2022 at 17:58
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    $\begingroup$ Also, $2A_{4k}$ contains the central product of $k$ copies of the quaternion group $Q_8$ so its minimum complex degree is at least $2^k$. $\endgroup$ Jan 20, 2022 at 22:42
  • $\begingroup$ @RichardLyons: Thanks- much better bound in that case! $\endgroup$ Jan 21, 2022 at 9:46

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