The commutator of two subgroup in a finite group

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Let $G=P\ltimes Q$ be a non-nilpotent group, where $P$ is a $p$-group and $Q$ is a cyclic $q$-group with distinct prime numbers $p,q$. Set $H:= \Omega_1(Q)$. Then why $[H,P] \neq 1$?

${\bf {My Try}}$: Since $Q$ is cyclic, $|H|=q$. Thus since $H \trianglelefteq G$, we have $[H,P] \leq H$, and so $|[H,P]|= 1$ or $q$. On the other hand we have $H=[H,P]C_H(P)$. Now if $[H,P]=1$, the $H=C_H(P)$ that is $H \leq C_G(P)$.