If $|G| = pq$, then $G$ has an element of order $q$

Question

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.

Answer 1

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.

Answer 2

You are missing an argument for why you have some element not of order $p$ (apart from the neutral element). Let's first deal with the case where $a$ generates a normal subgroup. In this case, let $b$ be some element not in the subgroup generated by $a$. Now $\langle a\rangle\langle b\rangle$ is a subgroup (since $\langle a\rangle$ was assumed normal), and it is easy to see that if $b$ has order $p$ then that subgroup has order $p^2$ which is not possible by Lagrange. So now either $b$ generates all of $G$ in which case we are done, or $b$ has order $q$ and we are again done.

Now assume that $\langle a\rangle$ is not normal and look at the set $K = \{g\langle a\rangle g^{-1}|g\in G\}$. It is a nice exercise to show that $|K| = |G:N_G(\langle a\rangle)|$ ($N_G( \langle a \rangle )$ is the normalizer of $\langle a\rangle $ in $G$). Since $\langle a\rangle $ is not normal the only possibility is that we then have $|K| = q$ (by Lagrange). Now, the subgroups $g\langle a\rangle g^{-1}$ are either identical or intersect only in the neutral element (again by Lagrange), so the total number of elements in these subgroups is $(p-1)q + 1$. This might not be all elements of the group, but if all elements of $G$ have order $p$ (apart from the neutral element), then we can pick out one more of those and repeat and we get that $pq = (p-1)qn + 1$. But this is impossible as $q$ divides the left side but not the right side.

I realize this is rather long, but it was the shortest proof I could produce that did not use anything but Lagrange (and even then I did have to leave a bit as an exercise).

Answer 3

I'll give you an alternative proof. This is a direct corollary of Cauchy's theorem.

It states the following: if G is a finite group and $p$ is a prime which divides #G, then $G$ contains an element of order $p$.

Proof:

$\#G$ = $p^rm$ with $r \geq 1$ and $p$ does not divide $m$. Now we can apply the first sylow theorem, which states that there is a subgroup of order $p$. Let us call this subgroup $H$.

Now take $x\in H$ with $x \neq e_H$, hence $order(x) \neq 1$. Now Lagrange's theorem tells us that $order(x) = \#\langle x\rangle$ divides $\#H$, but $p$ is prime, thus $order(x) = p$.