So I was given the fact that the Monster Group is a non abelian group of order $2^{46} · 3^{20} · 5^9 · 7^6 · 11^2 · 13^3 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71$.

But what I do not quite understand is how to show that there is a subgroup of this Monster isomorphic to $\mathbb{Z}$ or $\mathbb{Z_2}$.

Any suggestions will be helpful and much appreciated.

  • $\begingroup$ A subgroup isomorphic to $\mathbb{Z}?$ If am not wrong this is not possible. $\endgroup$
    – mfl
    Nov 10 '16 at 14:16
  • $\begingroup$ Could you clarify what "large" order means, as you are using it? The term "large" is relative. $\endgroup$
    – amWhy
    Nov 10 '16 at 14:24
  • $\begingroup$ By large I just meant that it is a Large number. (Did not want to type it out the first time around.) $\endgroup$ Nov 10 '16 at 14:26
  • $\begingroup$ Please check that I TeXified the order correctly. Surely the list of factors was supposed to be powers of primes. Also, the question is still non-sensical given that the Monster is finite. $\endgroup$ Nov 10 '16 at 14:28
  • 1
    $\begingroup$ In that case I hazard a guess that the intended solution is to imitate the argument here. See e.g. Arturo Magidin's answer. $\endgroup$ Nov 10 '16 at 14:33

Hint: Any finite group of even order contains an element of order $2$, see here, or use Cauchy's theorem.


Hint: Consider two cases, one in which the order is finite, one for which the order is infinite, recalling that any element generates a cyclic group whose size is equal to the order of the element.

  • $\begingroup$ Order of what? I know the order of the Monster Group is large. Am I supposed to find a subgroup of finite order and then a subgroup of infinite order from the Monster Group? $\endgroup$ Nov 10 '16 at 14:13
  • $\begingroup$ Order of the Monster Group, yes. $\endgroup$
    – b00n heT
    Nov 10 '16 at 14:14
  • 3
    $\begingroup$ Are you aware that the monster group is finite? $\endgroup$ Nov 10 '16 at 14:20
  • $\begingroup$ @TobiasKildetoft yes, I am aware of it, that's why I sense that trying to produce an subgroup isomorphic to $\mathbb{Z}$ is impossible, but I am not completely convinced that this is so for $\mathbb{Z_2}$ $\endgroup$ Nov 10 '16 at 14:30
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    $\begingroup$ Quarternion: I think @Tobias was directing his comment to address this answer and its author, and not you! $\endgroup$
    – amWhy
    Nov 10 '16 at 14:39

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