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

By elementary techniques (I mean, just by Sylow theory and without transfer theory), it can be shown that $G$ has normal Sylow-$p$ subgroup, call it $P$.

Then $|G/P|=qr$ is relatively prime with $|P|$.

By Schur-Zassenhaus theorem, there exists a subgroup $K$ of order $qr$ in $G$ such that $P\cap K=1$ and $PK=G$ (i.e. $K$ is a complement of $P$ in $G$).

Question: Without Schur-Zassenhaus theorem, can we show that $G$ contains subgroup of order $qr$?


Define : $G/P=:L$ the quotient group and $\pi$ the projection of $G$ onto $L$. We know (using semi-direct product and Sylow-theory arguments) that we will always have :


Here $\mathbb{Z}_n$ denotes the cyclic group of order $n$. In particular $\mathbb{Z}_q\triangleleft L$ and hence :

$$\pi^{-1}(\mathbb{Z}_q)\triangleleft G $$

Now, by standard enumerative arguments $\pi^{-1}(\mathbb{Z}_q)$ has cardinal $Ker(\pi)\times q=pq$. Let us denote this group $H$. We know that $H$ has for complement in $G$ any $r$-Sylow. Let us denote $R$ one of them. Furthermore, since $H$ is a group of order $pq$ we know that we actually have (where $Q$ is a $q$-Sylow of $G$) :

$$H=P\rtimes_{\psi_1}Q $$

Whatever might be $\psi_1$ we know that we will have $1$ or $p$ $q$-Sylows in $H$. Since $H$ is normal, we will have $1$ or $p$ $q$-Sylows in $G$ as well (show that any $q$-Sylow of $G$ is actually in $H$).

In other words $n_q$ cannot be divisible by $r$, hence $r$ divides the normalizer of $Q$, this implies that the group generated by $Q$ and $R$ is of order $qr$ and it is your complement.


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