Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@Kannappan: OP probably wants $p$ and $q$ prime methinks. Mark: Why must $b$ be of different order? – anon Jan 20 '12 at 8:12
Do you know the Sylow Theorems? – Lennart Jan 20 '12 at 8:24
Sylow's Theorems? I don't think you'll need Sylow's Theorems even to classify all groups of order $pq$, where $p,q$ are primes, not necessarily distinct. – user21436 Jan 20 '12 at 8:29
Let me elaborate on anon's comment: This is a gap you cannot close. Consider the symmetry group of a regular $q$-gon $D_q$, where $q > 2$ is prime. We have $|D_q| = 2 q~$ (and now set $p = 2$ to see what happens more clearly). Any reflection is of order $2 = p$. Let $a$ be any reflection and $b$ any other reflection. Then $b$ is not an element of the subgroup generated by $a$ but is also of order $2$, showing that your argument flawed. – Lennart Jan 20 '12 at 11:46
up vote 5 down vote accepted

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.

share|cite|improve this answer
Why is $H$ normal in $G$? – Tobias Kildetoft Jan 20 '12 at 21:46
@Tobias Yikes! This is bogus if we're not abelian. Thanks for pointing that out. No idea what I was thinking with this. – Dylan Moreland Jan 20 '12 at 21:56
Actually, I can fix the old proof up, but it's longer now. Oh well. – Dylan Moreland Jan 20 '12 at 22:23
Dear Dylan: I don't understand the last sentence of your answer. I would also have accepted it. – Pierre-Yves Gaillard Jan 21 '12 at 4:14

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).

share|cite|improve this answer
MathJax/LaTeX tip: use \langle and \rangle instead of < and >; the spacing for the latter is bad. – Arturo Magidin Jan 20 '12 at 16:19
Thanks for editing my lazy brackets. – Tobias Kildetoft Jan 20 '12 at 21:47

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 $p$.


$\#G$ = $p^rm$ 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 $order(x) = p$.

share|cite|improve this answer
I do not know sylow theorem and I am not sure if it's necessary to use it. Do you have an approach that only uses lagrange's theorem? – Mark Jan 20 '12 at 9:04
I added the condition that they are distinct primes. – Mark Jan 20 '12 at 9:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.