on the commutator subgroup Let $G=P \ltimes Q$ be a non-nilpotent group, where $P$ is a non-cyclic $p$-group and $Q$ is a cyclic $q$-group such that $|Q| \geq q^2$, ($p,q$ are two distinct primes). Then why $[P, \Omega_1(Q)] \neq 1$?
My Try: Since $\Omega_1(Q)$ char $Q \trianglelefteq G$, we get $\Omega_1(Q) \trianglelefteq G$. Hence $[P, \Omega_1(Q)] \leq \Omega_1(Q)$. Since $Q$ is cyclic, $\Omega_1(Q) \cong \Bbb{Z}_q$, and so $[P, \Omega_1(Q)]=1$ or $[P, \Omega_1(Q)]=\Omega_1(Q)$.

Let $Q = \langle x \rangle$ where $x$ has order $q^n$ with $n \ge 1$. Since $G$ is not nilpotent, $P$ does not centralize $Q$, so there exists $y \in P$ that does not centralize $x$.
So $y^{-1}xy = x^i$ with $i \ne 1$. We want to prove that $y$ does not centralize $\Omega_1(Q) = \langle x^{q^{n-1}}\rangle$.
If it did, then we would have $q^{n-1}i \equiv q^{n-1} \bmod q^n$, so $i \equiv 1 \bmod q$, i.e. $i = 1+kq$ for some $k$. Then $i^{q^{n-1}} \equiv 1 \bmod q^n$, which mean that the order of the automorphism of $Q$ induced by conjugation by $y$ divides $q^{n-1}$. But the order of $y$ is a power of $p$ with $p \ne q$, so this is impossible.
So $y$ does not centralize $\Omega_1(Q)$ and hence $[P,\Omega_1(Q)] = \Omega_1(Q)$.
There is no need to assume that $P$ is not cyclic, and there is no need to assume that $|Q| \ge q^2$.
