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I am doing the following exercise from [Birkhoff and MacLane, A survey of modern algebra]:

Let $G$ be a group of order $pq$ ($p,q$ primes). Show that either $G$ is cyclic or contains an element of order $p$ (or $q$). In the second case, show that $G$ contains either $1$ normal or $q$ conjugate subgroups of order $p$. In the latter case the $pq-(p-1)q=q$ elements not of order $p$ form a normal subgroup. Infer that $G$ always has a proper normal subgroup.

I think I was able to prove $G$ has a proper normal subgroup, but without quite establishing that the number of subgroups of order $p$ is 1 or $q$.

Here is my solution:

Suppose $G$ is cyclic and $G=\langle x \rangle$. Then, $\langle x^p \rangle$ has order $q$ and is normal, so we are done. So suppose $G$ is not cyclic. Let $x \in G-1$. By Lagrange's theorem, the order of $x$ is $p$, say. Let $H:=\langle x \rangle$. $G$ acts on all subgroups of order $p$ by conjugation. In this action, the stabilizer of $H$ is the normalizer $N_G(H)$. Since $N_G(H) \supseteq H$, $|N_G(H)|$ equals $p$ or $qp$. Hence the orbit containing $H$ has length $q$ or 1 (by the orbit-stabilizer lemma). If this value is 1, then $H$ is normal, so we are done. If this value is $q$, then there are $pq-(p-1)q=q$ remaining nonidentity elements in $G$ that are not in any conjugate of $H$.

Case 1: If any of these $q$ elements, say some $y \in G$, has order $p$, then $K:=\langle y \rangle$ is either normal (so we are done) or has $q$ conjugate subgroups (in which case the conjugates of $H$ and of $K$ together contain $q(p-1)+q(p-1)+1>qp$ elements, a contradiction).

Case 2: If none of the $q$ elements has order $p$, then they all have order 1 or $q$, and hence form a cyclic subgroup $K$ of order $q$. $K$ must be normal, since otherwise $g^{-1}Kg \ne K$ for some $g \in G$, whereas $K$ already exhausted all the elements of $G$ of order $q$.

Thus, $G$ has a normal subgroup of order $p$ or $q$.

While I have shown the group is not simple, I think I haven't shown yet that the number of subgroups of order $p$ is 1 or $q$. The proof only establishes that the number of distinct conjugates of $H$ is 1 or $q$. Any suggestions on how to prove (using elementary methods) that any other subgroup of order $p$ would have to be conjugate to $H$? In other words, I need to rule out Case 1 in the proof.

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    $\begingroup$ I think this follows more easily by the Sylow theorems. Can you use them? $\endgroup$
    – Ian Coley
    Feb 6, 2014 at 17:57
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    $\begingroup$ I know the Sylow conjugacy theorem which would establish the remaining part I need to prove, but I would like some kind of elementary proof or actually some kind of modification of my existing proof. I mean, can the proof above be "fixed" to rule out Case 1? $\endgroup$
    – AG.
    Feb 6, 2014 at 17:59
  • $\begingroup$ Well, by Cauchy theorem there are elements of order p and q. So there, you can actually stop since $p \lor q$ is true whenever one of the two is true. $\endgroup$
    – user370967
    Jul 24, 2017 at 13:02

4 Answers 4

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In your proof (in the case that $G$ is not cyclic), you get a normal subgroup $K$. Say it has order $q$. Let $H$ be a subgroup of order $p$. If $H$ is normal in $G$, then $G$ will be cyclic. We get that $H$ has $q$ conjugates. But now we conclude that $q+q(p-1)=|G|$ elements in $K $ and the conjugate of $H$. Hence there is no other subgroup of order $p$ except the conjugate of $H$. (All the proof holds under the condition $p \ne q$).

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  • $\begingroup$ I think your answer refers to Case 2 of my proof, which is already settled. My question is on how to rule out Case 1. $\endgroup$
    – AG.
    Feb 7, 2014 at 2:48
  • $\begingroup$ How about Cauchy theorem $\endgroup$
    – Wei Zhou
    Feb 7, 2014 at 3:21
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  1. You say "If this value is $q$ then there are $pq-(p-1)q=q$ remaining nonidentity elements in $G$..." Not quite. There are $q$ remaining elements. One of the remaining elements is the identity $1$.

Here are a few details that are implicit in your argument (probably too easy to need to be spelled out, but, better safe than sorry):

First, if $H$ is any subgroup with $p$ elements, note that every nonidentity element of $H$ is a generator for $H$.

If $H_1$ and $H_2$ are any two subgroups with $p$ elements each, if $H_1 \cap H_2$ contains any nonidentity element, then that element in the intersection is a generator for both $H_1$ and $H_2$. So $H_1=H_2$. Therefore, if $H_1$ and $H_2$ are any distinct $p$-element subgroups, then they are also disjoint (except for the identity element).

This is the reason why the union of the $q$ conjugate groups of $H$ contains $q(p-1)$ nonidentity elements.

  1. You want to rule out Case 1. It seems to me that your argument does this. Can you say what part worries you? If there is an element $y$ which is not in any conjugate of $H$, and $y$ has order $p$, then $K = \langle y \rangle$ is another subgroup of order $p$, and the conjugates of $K$ again contain $q(p-1)$ nonidentity elements. No conjugate of $K$ shares any nonidentity element with any conjugate of $H$, otherwise $K$ and $H$ would be conjugate. So all together these conjugates of $K$ and $H$ have $2q(p-1)$ nonidentity elements, or $2q(p-1)+1$ elements in total. But $2q(p-1)+1=pq+q(p-2)+1>pq$, as you noted. This is impossible, the union of the conjugates can't have more elements than $G$. So the element $y$ can't have order $p$. This shows that every element $y$ outside the conjugates of $H$ has order $q$.

Since $q$ is prime, every nonidentity element $K = \langle y \rangle$ has order $q$. This accounts for all the $q-1$ elements of order $q$. So, this set of $q$ elements left over from the conjugates of $H$ must be equal to $K$.

And $K$ must be normal because any conjugate of $K$ consists of order $q$ elements, but all the order $q$ elements are already in $K$.

In conclusion, your argument seems fine (modulo a minor wording thing in point (1) above), and in particular it seems to me that you already ruled out Case 1.

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Hints:- Prove at first the following lemma by Extended Cayley's Theorem

If |G|=$ pn $ where p>n (p is a prime) then G has a normal subgroup H of order p.

Now let p>q then by the above lemma G has a normal subgroup of order p and vice versa if q>p. If p=q then Z(G) is the desired normal subgroup of order p since |G|= $ p^2 $. Hence the proof.

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Let $G$ be a group of order $pq$ ($p,q$ primes). Show that either $G$ is cyclic or contains an element of order $p$ (or $q$).

The nontrivial elements of $G$ can have order $p$ or $q$ or $pq$, only (Lagrange theorem). If $G$ is noncyclic, then the third option is ruled out.

In the second case, show that $G$ contains either $1$ normal or $q$ conjugate subgroups of order $p$.

Say $k_p$ and $k_q$ the number of elements of order $p$ and $q$, respectively. They build up $n_p$ and $n_q$ cyclic subgroups, respectively, which are mutually trivially intersecting (Lagrange). Therefore, $k_p=n_p(p-1)$ and $k_q=n_q(q-1)$, and hence $pq=$ $1+k_p+k_q=$ $1+n_p(p-1)+n_q(q-1)$. This latter has solutions $(n_p,n_q)=$ $(1,p),(q,1)$; in both solutions, the non-unique subgroups are conjugate, because otherwise $G$ would act non-transitively (by conjugation) on a set of size $p$ or $q$, requiring $|G|$ to have factors of them (the size of an orbit divides the order of the group), a contradiction because they are both primes. This also proves the final part:

In the latter case the $pq-(p-1)q=q$ elements not of order $p$ form a normal subgroup. Infer that $G$ always has a proper normal subgroup.

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