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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 $7$.

Sebastien Palcoux
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1 Answers1

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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).

Geoff Robinson
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  • 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? – Sebastien Palcoux Jan 15 '20 at 16:11
  • 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. – Geoff Robinson Jan 15 '20 at 16:24
  • I meant a weaker assumption on the decomposition:

    • Firstly, perhaps something like: the existence of (at least) one irrep of dim $5$ and of dim $7$ and $$5_1 \otimes 5_1 \ge 1 \oplus 5_1 \oplus 7_1$$ would be enough for your argument, correct?

    • And secondly, perhaps the couple $(5,7)$ can be generalized to a class of couples $(n,m)$, isn't it?

     – Sebastien Palcoux Jan 15 '20 at 16:46
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    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. – Geoff Robinson Jan 15 '20 at 18:27
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    @Geoff: Why is it true (in your notation) that $G/K$ has the same property as $G$? –  Jan 16 '20 at 02:12
  • @MarkSapir: a fusion subcategory of $\mathrm{Rep}(G)$ is isomorphic to $\mathrm{Rep}(G/N)$ with $N \unlhd G$ (and reciprocally). – Sebastien Palcoux Jan 16 '20 at 08:30
  • @SebastienPalcoux: What if $N=G$? –  Jan 16 '20 at 08:35
  • @MarkSapir: $N=G$ corresponds to the trivial subcategory (i.e. given by the trivial irrep). – Sebastien Palcoux Jan 16 '20 at 08:37
  • @MarkSapir : Because if $5_{1}$ has $K$ in its kernel, so does every representation on the right side, because $5_{1} \otimes 5_{1}$ has $K$ in its kernel. – Geoff Robinson Jan 16 '20 at 08:55
  • @SebastianPalcoux: I do not know what fusion sucategory is. But whatever it is it cannot be the same for $G$ and for $G/N$ if $N=G$. An easier question is (in Geoff's notation) why $G/K$ has an irrep of degree 7? If this is true it should have something to do with the fact that $K$ is the kernel of nn irrep of degree 5. It can't be true for arbitrqry normal $K$. –  Jan 16 '20 at 09:03
  • @MarkSapir: the isomorphic classes of irreps form a ring under the tensor product (it has the strucutre of a fusion ring, and it is the Grothendieck ring of the fusion category $\mathrm{Rep}(G)$, but it does not matter). Now let consider $5_1$ as an element of this fusion ring, it generates a fusion subring which, as every fusion subring, is isomorphic to the Grothendieck ring of $\mathrm{Rep}(G/N)$ with $N$ a normal subgroup depending on the subring (here $N=ker(5_1)$). – Sebastien Palcoux Jan 16 '20 at 09:40
  • @MarkSapir: Dietmar and Vaughan know all that, so if you and (one of) them are currently at Vanderbilt, it could be a good opportunity to discuss. – Sebastien Palcoux Jan 16 '20 at 09:49
  • @MarkSapir: As I explained above, it is because the normal subgroup $K$ is the kernel of $5_{1}$, the statement would not be true in general for an arbitrary normal subgroup. Once $K$ is in thhe kernek of $5_{1}$, $K$ is also in the kernel of everything in sight in the original equation, so, as I said, $G/K$ has the same property (regarding $5_{1}$ as a representation of $G/K)$. – Geoff Robinson Jan 16 '20 at 10:20
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    @Geoff: Thabk you for the explanation. –  Jan 16 '20 at 12:34