TITLE: If $|G| = pq$, then $G$ has an element of order $q$
QUESTION [6 upvotes]: Suppose I have a group $G$ order $pq$, where $p$ and $q$ are distinct primes, and I know that there exists $a \in G$ such that $order(a) = p$. How do I show that there exists $b \in G$ such that $order(b) = q$?
I have tried by arguing that since $a$ is order $p$ then there must be a element that isn't generated by $a$, call it $b$, then $b$ must be of different order, if $order(b) =pq$ then by corollary of Lagrange's theorem $b^{p}$ is of order q. Otherwise the only choice left is $order(b) = q$.
I think there is a flaw in my approach. Please point out any errors.
Notes: I forgot to mention that pq are distinct primes.

REPLY [5 votes]: As Dimitri says, it is enough to show that if $G$ is a finite group and $p$ a prime dividing the order of $G$, then $G$ contains a subgroup of order $p$. This is known as Cauchy's theorem. Here's a slick proof of this that I found in an old prelim. It appears to be due to McKay.
Let
$$
S = \{(x_1, \ldots, x_p) \in G^p \mid x_1 \cdots x_n = e\}.
$$
This has order $|G|^{p - 1}$. The cyclic group $C_p$ of order $p$ acts on this set by shifts: a generator sends $(x_1, \ldots, x_p)$ to $(x_p, x_1, \ldots, x_{p - 1})$, and so on. The orbits of this action have size $1$ or $p$, and as the orbits partition $S$ and $p \mid |S|$, it follows that the number of fixed points is zero modulo $p$. But there is at least one fixed point, namely $(e, \ldots, e)$. So there exists an element $(x, \ldots, x)$ in $S$ with $x \neq e$, and this is what you want.
I had an attractive proof of this fact which was also clearly false (thanks to Tobias for pointing that out). Fixing it up would require almost as much work as showing the existence of Sylow subgroups, so I've added an alternative proof. This certainly shouldn't be the accepted answer, however.