# Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation

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 $$7$$.

• 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. – Nick Gill Jan 15 '20 at 14:56
• @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. – Sebastien Palcoux Jan 15 '20 at 15:00
• @NickGill Confirmation that there is no such simple group by the paper of Hiss-Malle Low-dimensional Representations of Quasi-simple Groups and corrigenda. – Sebastien Palcoux Jan 15 '20 at 15:32
• Ah, yes, I was in error about perfect groups, but glad to hear that is confirmed for simples. – Nick Gill Jan 15 '20 at 15:46

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).

• 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 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
• @Geoff: Why is it true (in your notation) that $G/K$ has the same property as $G$? – user6976 Jan 16 '20 at 2:12
• @Geoff: Thabk you for the explanation. – user6976 Jan 16 '20 at 12:34