The last three prime numbers in the factorization of the order of the Fischer Griess friendly monster group are 47, 59, 71. (https://en.wikipedia.org/wiki/Monster_group) On the other hand, the monster group can be contrcuted from the automorphism group of the Griess algebra which has 196884dimensions. The Monster fixes (vectorwise) a 1space in this algebra and acts absolutely irreducibly on the 196883dimensional orthogonal complement of this 1space (https://en.wikipedia.org/wiki/Griess_algebra). The factorization of 196883 is 47*59*71. Is this concidence or can it be understood from the construction of the group?
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3$\begingroup$ The first rule of moonshine is that there are no coincidences. That said, if you look at the dimensions of smallest nontrivial representations of other sporadic groups, you also get the largest prime factor or two in the order. For example, $4371 = 3*31*47$, and the largest prime factors in the order of the baby monster are 31 and 47. Anyway, the fact that the dimension of an irreducible representation must divide the order of the group should cut down on the surprise a bit. $\endgroup$ – S. Carnahan♦ Jul 17 '16 at 7:02

$\begingroup$ If you look at all $194$ degrees of the irreducible representations of $\mathbb{M}$ as A001379 sequence $1, 196883, 21296876, 842609326, \dots$, then $$\begin{aligned} 196883 &= 47\times 59\times 71\\ 21296876 &= 2^2\times 31\times 41\times 59\times 71\\ 842609326 &= 2\times 13^2\times 29\times 31\times 47\times 59 \end{aligned}$$ and so on, with precisely the same prime factors also dividing the order of $\mathbb{M}$. $\endgroup$ – Tito Piezas III Mar 31 '18 at 11:37

$\begingroup$ @S.Carnahan: There is also the coincidence $163+67+43+19+11+7 = 310$ where $310$ is the number of objects involving certain moonshine groups. Kindly see this post. $\endgroup$ – Tito Piezas III Apr 4 '18 at 12:03
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