# 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. Jan 15, 2020 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. Jan 15, 2020 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. Jan 15, 2020 at 15:32
• Ah, yes, I was in error about perfect groups, but glad to hear that is confirmed for simples. Jan 15, 2020 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? Jan 15, 2020 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. Jan 15, 2020 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. Jan 15, 2020 at 18:27
• @Geoff: Why is it true (in your notation) that $G/K$ has the same property as $G$?
– user6976
Jan 16, 2020 at 2:12
• @Geoff: Thabk you for the explanation.
– user6976
Jan 16, 2020 at 12:34