# Isomorphisms of the Monster Group

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.

• 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. 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.) 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. Nov 10 '16 at 14:28
• In that case I hazard a guess that the intended solution is to imitate the argument here. See e.g. Arturo Magidin's answer. 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.
• @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 $\mathbb{Z_2}$ Nov 10 '16 at 14:30