The dimensions of some representations of the Janko group J1 coinside with dimensions of smallest representations of the Lie algebra of type E7 (56, 133). It seems to be natural that there is a subgroup of the Lie group of type E7 isomorphic to J1. Is it true?
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2$\begingroup$ If there is an embedding J1 --> E7, then the next smallest representations of E7 should also decompose into (hopefully few) J1-representations. Have you checked whether that is indeed the case? $\endgroup$– André HenriquesCommented Jun 2, 2018 at 20:18
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3$\begingroup$ For the dimension of next smalles representation we have: 912=209+3*133+4*76. $\endgroup$– Ievgen MakedonskyiCommented Jun 2, 2018 at 20:35
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8$\begingroup$ J1 is a subgroup of a reduction of G2 mod 11. Possibly one can find a copy of G2 in E7 and show that the resulting linear representations mod 11 lift to characteristic zero. (This guess is motivated by the known observation that Thompson's sporadic group, though not contained in E8(C), is a subgroup of a reduction mod 3 of E8 and has a 248-dimensional representation over C.) $\endgroup$– Noam D. ElkiesCommented Jun 2, 2018 at 21:22
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2$\begingroup$ Another obvious check: if J1 embeds into E7, the 56-dimensional representation must carry invariant 4-form, so the fourth symmetric power must contain a trivial representation. I guess it can be checked via the character table and, if not, the conjecture is false, and if yes, you can try to catch this 4-form and compare it to that for E7. $\endgroup$– Victor PetrovCommented Jun 6, 2018 at 20:05
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2$\begingroup$ The finite simple groups (and their central extensions) that embed into exceptional Lie groups were classified across many papers. There is a summary by Griess and Ryba here. J1 does not appear to be one of them. But incidentally following Victor's suggestion I got Sym^4(56)^{J1} is 8-dimensional, for both the 56-dimensional representations. $\endgroup$– David TreumannCommented Jun 9, 2018 at 20:48
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1 Answer
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The Frobenius–Schur indicators of both $56$-dimensional representations are $+1$, meaning that in both cases the representation supports an invariant symmetric bilinear form. Since these are irreps, they do not support any other invariant bilinear forms. On the other hand, the $56$-dimensional irrep of $\mathrm{E}_7$ supports an invariant antisymmetric bilinear form. So $\mathrm{J}_1 \not \subset \mathrm{E}_7$.
According to the ATLAS, the finite group of Lie type $\mathrm{E}_7(2)$ has a $132$-dimensional "adjoint" representation over $\mathbf{F}_2$. The ATLAS does not list such a representation for $\mathrm{J}_1$, but I don't know how to confirm that it does not exist, so I cannot rule out an inclusion $\mathrm{J}_1 \overset?\subset \mathrm{E}_7(2)$.
I remark that of the three $133$-dimensional representations of $\mathrm{J}_1$, the first supports one invariant antisymmetric 3-form, which could in principle be a Lie bracket, and the other two support five each.
answered Aug 29, 2019 at 1:27
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5$\begingroup$ By the main theorem of "Kleidman, Peter B.; Wilson, Robert A. Sporadic simple subgroups of finite exceptional groups of Lie type. J. Algebra 157 (1993), no. 2, 316–330.", the sporadic group $J_1$ is a subgroup of $E_7(q)$ if and only if $q = 11^a$. $\endgroup$ Commented Aug 29, 2019 at 13:31
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$\begingroup$ @MikkoKorhonen I cannot speak for the original poster, but I suggest you post that comment as an answer, because it deserves to be accepted. $\endgroup$ Commented Aug 31, 2019 at 4:00