TITLE: Prove that a group of order 42 must have a subgroup of order 6
QUESTION [5 upvotes]: Prove that a group of order 42 must have a subgroup of order 6.

Firstly, I use Sylow's theorem to show there must exist a subgroup of order 7. What about 6?

REPLY [3 votes]: The number of Sylow 3-subgroups:

is congruent to $1\pmod{3}$
divides $|G|$
is equal to $[G:N_G(P)]$ for any Sylow 3-subgroup $P$

Points (1) and (2) show that the number of Sylow 3-subgroups is either $1$ or $7$.
If there's only one, it's normal (and we can talk about the Sylow 3-subgroup $P$).  Then if $Q$ is a Sylow 2-subgroup, $PQ$ is a subgroup (because $P$ is normal), and it has order $6$.
If there are seven Sylow 3-subgroups, then point (3) shows $|N_G(P)|=6$, and we're already done.