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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.

rubik
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Quarternion
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  • A subgroup isomorphic to $\mathbb{Z}?$ If am not wrong this is not possible. – mfl Nov 10 '16 at 14:16
  • Could you clarify what "large" order means, as you are using it? The term "large" is relative. – amWhy Nov 10 '16 at 14:24
  • By large I just meant that it is a Large number. (Did not want to type it out the first time around.) – Quarternion Nov 10 '16 at 14:26
  • 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. – Jyrki Lahtonen Nov 10 '16 at 14:28
  • To give a helpful answer we need to know what you know about groups? If this is your first course in groups, then the question makes sense as an exercise from a slightly naughty teacher. If you are more than a few months into studying groups, then the question does not IMO make any sense. – Jyrki Lahtonen Nov 10 '16 at 14:31
  • @JyrkiLahtonen This is my first course in Group Theory. – Quarternion Nov 10 '16 at 14:32
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    In that case I hazard a guess that the intended solution is to imitate the argument here. See e.g. Arturo Magidin's answer. – Jyrki Lahtonen Nov 10 '16 at 14:33
  • I am tempted to close this as a duplicate of one of the linked questions, because that same argument works for the Monster. Vote/comment, please. – Jyrki Lahtonen Nov 10 '16 at 14:39
  • @JyrkiLahtonen Yes, you are right. The question really has nothing to do with the monster. Any finite group of even order will do, and your link is the right one (I saw it only after my answer, I am sorry). So I don't mind closing. – Dietrich Burde Nov 10 '16 at 19:24

2 Answers2

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

Community
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Dietrich Burde
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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.

b00n heT
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  • 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? – Quarternion Nov 10 '16 at 14:13
  • Order of the Monster Group, yes. – b00n heT Nov 10 '16 at 14:14
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    Are you aware that the monster group is finite? – Tobias Kildetoft Nov 10 '16 at 14:20
  • @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 $\mathbb{Z_2}$ – Quarternion Nov 10 '16 at 14:30
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    Quarternion: I think @Tobias was directing his comment to address this answer and its author, and not you! – amWhy Nov 10 '16 at 14:39