on the existence of a subgroup

Question

Let $G$ be a non-nilpotent group such that $p,q \in \pi(G)$, the set of all prime divisor of $|G|$. If $p \mid q-1$, then can we say that there is subgroup $S$ which has the form $S=P_1 \ltimes Q_1$, where $|Q_1|=q$ and $|P_1|=p^t$?

Answer 1

As ancientmathematician pointed out, this isn't true in general. However, there is a smaller, easier to comprehend counterexample.

Consider the group of order 72 given by $D_8\ltimes 3^2$ with $D_8$ acting by the usual matrices of 0s, 1s, and -1s that also represent it over the real plane. There is no normal subgroup of order 3. (For clarity, $3^2$ is the standard group-theory notation for the elementary abelian group of order 9.)

Indeed, if your statement were true even for solvable groups, then all irreducible representations of all $p$-groups over $\Bbb{F}_q$ would be one-dimensional, which is clearly not the case.

Answer 2

Consider $G=\text{SL}(2,64)$, $p=3$, $q=13=1+4\cdot3$.

The normaliser of the Sylow-$13$-subgroup is $D_{2\cdot65}$.