If $l,p,q$ are primes with $l<p<q$, such that

$$p\nmid (q-1)\hspace{1cm} l\mid (p-1)\hspace{1cm} l\mid (q-1) $$

I want to show that there are at least $1$ and at most $(l+1)$ non-abelian groups of order $lpq$.

I already showed (using Sylow theorems) that in this case, if $G$ is a group with order $lpq$ then

$$ G \cong C_{pq}\rtimes C_l .$$

The fact that there is at least one non-abelian group I showed the following way: we know that $|\text{Aut}(C_{pq})| = (p-1)(q-1)$, so $l$ divides its order, and as a consequence of the Sylow theorems we know that $C_l\hookrightarrow \text{Aut}(C_{pq})$, thus there is a non-trivial homomorphism which gives us a non-abelian group $C_{pq}\rtimes C_l$.

The problem I'm having is showing that $\text{Aut}(C_{pq})$ can have at most $(l+1)$ copies of $C_l$ in it. Any hints?


An answer to your question is provided here (page 73):


You'll see that if $l$, $p$, $q$ satisfy your conditions then there are always $p+1$ non-abelian groups of order $lpq$.

Another reference for that fact is the book by Blackburn, Neumann, and Vekataraman "Enumeration of finite groups", page 238.

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  • $\begingroup$ Thank you Very much! $\endgroup$ – D18938394 May 4 '18 at 2:20
  • $\begingroup$ Have you been able to understand what's written there? $\endgroup$ – the_fox May 5 '18 at 17:41

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