In this post, irrep and dim mean "irreducible complex representation" and "dimension", respectively. It would be helpful (in a problem of monoidal category) to find a finite group $G$ with (at least) two irreps of dim $5$ (denoted $5_1$ and $5_2$) and (at least) two irreps of dim $7$ (denoted $7_1$ and $7_2$) with $$ 5_1 \otimes 5_1 \simeq 1 \oplus 5_1 \oplus 5_2 \oplus 7_1 \oplus 7_2$$Question: Is there such a finite group $G$?

Remark: such group should be of order a multiple of $35$ and should admit irreps of dims $5$ and $7$, which is helpful to rule out the following cases with GAP on a laptop:

  • simple, of order less than $10^6$,
  • perfect, of order less than $15120$,
  • general, of order less than $2240$.

Let $a_n$ be the smallest order of a group with an irrep of dim $n$ (oeis.org/A220470): $1, 6, 12, 20, 55, 42, 56, 72, 144, 110, 253, 156, 351, 336, 240, 272,\dots$

In particular, $a_5=55$ (given by $C_{11} : C_5$) and $a_7=56$ (given by $C_2^3 : C_7$). Now, I don't even know if there exists a group $G$ with $|G|<55 \times 56=3080$, and with irreps of dims $5$ and $7$.

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    $\begingroup$ I would think that if the presence of irreps of dim 5 and 7 is enough to rule out simple groups of order less than $10^6$, then it is enough to rule out ALL simple groups: larger simple groups will not have any irreducible representations that small (checking Landazuri-Seitz would be enough to confirm this). The same is probably true for perfect groups. $\endgroup$
    – Nick Gill
    Jan 15, 2020 at 14:56
  • $\begingroup$ @NickGill: the first perfect group with irreps of dims $5$ and $7$ is $A_5 \times \mathrm{PSL}(2,7)$ (of order $10080$) but it does not satisfy the expected decomposition. $\endgroup$ Jan 15, 2020 at 15:00
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    $\begingroup$ @NickGill Confirmation that there is no such simple group by the paper of Hiss-Malle Low-dimensional Representations of Quasi-simple Groups and corrigenda. $\endgroup$ Jan 15, 2020 at 15:32
  • $\begingroup$ Ah, yes, I was in error about perfect groups, but glad to hear that is confirmed for simples. $\endgroup$
    – Nick Gill
    Jan 15, 2020 at 15:46

1 Answer 1


I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful.

Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of $Z(G)$, then we must have $\lambda^{2} =1 $ since the trivial character occurs in $5_{1} \otimes 5_{1}$. However $\lambda = \lambda^{2}$ since $5_{1}$ also occcurs in $5_{1} \otimes 5_{1}$. Hence $\lambda$ is trivial.

Now $ N = O_{5^{\prime}}(G)$ is Abelian by Clifford's Theorem. If $N$ is non-trivial, then it is also non-central, sincee $Z(G) = 1$, and it follwws from Clifford's Theorem that the representation $5_{1}$ is (up to equivalence) monomial. Then $G$ has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{5}$. But in that case, $G$ has an Abelian normal Sylow $7$-subgroup, and $G$ has no irreducible character of degree $7$ (by a theorem of Ito, the degree of a complex irreducible character of $G$ divides $[G:A]$ whenever $A \lhd G$ is Abelian). Hence it follows that $N = 1$. More generally, this argument shows that the representation $5_{1}$ is primitive, ie not (equivalent to one) induced from any proper subgroup of $G$.

There are several ways to finish from here. One is to invoke Brauer's classification of the finite primitive subgroups of ${\rm GL}(5, \mathbb{C})$ and note that none of these has tthe order of $G/Z(G)$ divisible by $7$.

Another is to note that if $O_{5}(G)$ is non-trivial, then it is irreducibly represented by $5_{1}$, in which case $Z(G)$ has order divisible by $5$, a contradiction.

Now we are reduced to the case $F(G) = 1$ and we continue until we see that $M = F^{\ast}(G)$ is a finite simple subgroup of ${\rm GL}(5,\mathbb{C})$ of orer divisible by $35$. But by a theorem of Feit, if $G$ is a finite simple subgroup (of order divisible by the prime $p$) of ${\rm GL}(p-2,\mathbb{C})$ for some prime $p$, then $p$ is a Fermat prime and $G = {\rm SL}(2,p-1)$ ( we may apply this with $p = 7$, so obtain a contradiction).

  • $\begingroup$ My comment about Hiss-Malle is only for the simple groups. Your answer works for every group, right? Does your answer prove a more general statement than what expected? (because you seem to use just partially the assumption) If so, what is it? and to what could it be extended? $\endgroup$ Jan 15, 2020 at 16:11
  • $\begingroup$ Yes, this argument shows that there is no such finite $G$, simple or not. As for a more general statement, I am not sure: arguments about complex linear groups of low dimension tend to be rather ad hoc, and some arguments above are very specific to your hypotheses. $\endgroup$ Jan 15, 2020 at 16:24
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    $\begingroup$ I think it is probably true that if $p >3$ is a prime such that $q = p+2$ is also prime, then there is no finite group $G$ with a complex irreducible characters $\chi,\mu$ of respective degrees $p$ and $q$ such that $\chi^{2} = 1 + \chi + \mu + \theta$, where $\theta$ is a character ( or $0$). The argument (using Feit's Theorem rather than Brauer's) goes through more or less unchanged, after noting that $(3,5)$ is the only prime pair $(p,q)$ such that $p+1 = q-1$ is power of $2$. But this seems very specialized. $\endgroup$ Jan 15, 2020 at 18:27
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    $\begingroup$ @Geoff: Why is it true (in your notation) that $G/K$ has the same property as $G$? $\endgroup$
    – user6976
    Jan 16, 2020 at 2:12
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    $\begingroup$ @Geoff: Thabk you for the explanation. $\endgroup$
    – user6976
    Jan 16, 2020 at 12:34

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