If $q$ is not a divisor of $p-1$, then $G$ is cyclic. An open discussion on different proofs.

Question

Let $G$ be a group of order $pq$, where $p > q$ are primes.

a) If $q$ is not a divisor of $p-1$, then $G$ is cyclic.

I know the proof using Sylow's theorem but I was wondering if it can be proved without using Sylow's theorem. I always thought that even if such a proof will exist then that will be a proof of Sylow's theorem first without naming it because in I N Herstein this problem is given before discussing Sylow's theorem it propelled my thought again.

Here is my go what I think that everyone would do that Let $H$ be a subgroup of $G$ with order $p$ and $K$ be a subgroup of order $q$,

It is easy to see that $H$ must be normal in $G$

Now can we directly take the action of $K$ on $Aut(H)$ by conjugation without stating anything because this might give you one view point of generating groups. For a student facing the problem of I N Herstein might not know that this is the only way here!!

So what is your best bet here? Is there any way out? Say counting arguement or so!!

Because if we jot down the question the biggest non-trivial(please forgive me if it would be trivial for you then the rest would trivial as well, that is why..) fact is to prove that $K$ is normal. Can anyone prove it without using Sylow's theorem? Or any other interesting way is most welcome.

P.S: You can mark it as duplicate as a lot of similar question has been posted but let me justify why this post is different. The answers discussed there, are derived by using Sylow's theorem or Semi-direct product so I tried to have an open discussion on different proofs of the problem. Thanks a lot

Answer 1

Here's a purely numeric approach:

Let $h$ be a generator of $H$ (so $h^p = e$) and $k$ generate $K$ (so $k^q = e$). By normality of $H$, we have $k^{-1}hk = h^m$ for some $m$ in [1..p-1]. It is an easy induction from this to see that $k^{-i}hk^i = h^{m^i}$ for all $i$; from this we have specifically that $h^1 = h = k^{-q}hk^q = h^{m^q}$, so necessarily $1 = m^q$ (mod p). Also, by Fermat's Little Theorem, we have $1 = m^{p-1}$ (mod p). If $q$ does not divide $p-1$, then necessarily 1 = GCD($q$,$p-1$), so there are integers $r$, $s$ such that $1 = rq + s(p-1)$. Then $m = m^1 = m^{rq + s(p-1)} = (m^q)^r (m^{p-1})^s$ $= 1^r1^s = 1$ (mod p). Hence $k^{-1}hk = h$, so $h$ and $k$ commute, and so we have an (inner) direct product of $H$ and $K$ (which we see is cyclic, as $p$ and $q$ are relatively prime).

Answer 2

Consider just the morphism $G \rightarrow Aut(H)$ given by conjugation. Note that its range $R$ must divide both $|G|=qp$ and $|Aut(H)|=p-1$ (because $H$ cyclic, $|H|=p$).

So since $q$ and $p-1$ are coprime, $|R|=1$, ie the morphism is trivial, so the group is abelian and the conclusion follows easily.

Edit: proving that the group is abelian, given that the morphism is trivial, as mentioned in the comment, is not obvious.

Let $C$ be the subgroup of all elements commuting with $K$. $K$ is cyclic so $K \subset C$ so $q||C|$. Since the morphism above is trivial, $H \subset C$ thus $p||C|$ and thus $pq||C|$ ie $C=G$.

Thus the center $Z$ of $G$ contains both $H$ and $K$ so is $G$ (for similar reasons).