If a group is of order $m$ and $n$ divides $m$, must the group contain an element of order $n$?
Say we have a group of order $120$. Must the group contain an element with order $12$? If not, are there any theorems that tell us what order elements of a group must contain?
Comments:
If any prime $p$ divides the order of the group $|G|,$ then $G$ must have an element of order p. This is due to Cauchy's theorem. However, it is false if $p$ is not prime.
https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
Counterexample: Consider the group $Z_2 \times Z_2\times Z_2 \times Z_2.$ Clearly 4 divides the order of the group. Does this group have any element of order $4$?
No. If so then every group would be cyclic. For $m\mid m$. (Granted that's probably not what you meant.)
If we ask, instead, for $n\lt m$, the answer is still no.
An somewhat important example is $A_4$. It doesn't have a subgroup of index two. It's the smallest "non-Langrangian" group.
However, imho the best theorem to know of this flavor is Cauchy's theorem. It guarantees an element of order every prime divisor of $n$. My favorite proof is to use Sylow, to get a subgroup of order $p^l$. Then take the cyclic subgroup generated by any non-identity element. Cyclic groups have cyclic subgroups of every order dividing the order of the group. (There's a more basic proof, I am told, but I don't know it.)
There's also Sylow and Hall.
