I have a follow-up question to Problems in understanding a passage in the proof of Grün theorem for transfer. In the book of Kurzweil and Stellmacher, it is also concluded that
$G\neq O^p(G) \iff H\neq O^p(H)$.
Here $O^p(G)$ is the smallest normal subgroup with $p$-factor group.
My attempt: If $G=O^p(G)$, then $G$ has no proper normal subgroup with $p$-factor group. In particular, $G$ has no proper normal subgroup with $p$-factor Abelian group. So, $G= G'(p)$. Thus, $H= H'(p)$. Thus, $H$ has no proper normal subgroup with $p$-factor Abelian group.
However, to conclude, $H=O^p(H)$, I need that $H$ has no proper normal subgroup with $p$-factor group. How to show this?
Edit: The precise statement of the theorem is: Let $P$ be a Sylow $p$-subgroup of a finite group $G$ and $Z\leq Z(P)$ be weakly closed in $P$. Set $H:=N_G(Z)$. Then $P\cap G'=P\cap H'$ and $P/(P\cap G')\simeq G/G'(p) \simeq H/H'(p)$. In particular, $$G\neq O^p(G) \iff H\neq O^p(H).$$
My doubt is about proving the displayed equivalence.
Since $p$-groups are solvable, we have $$G/O^p(G)\ne 1\iff (G/O^p(G))'<G/O^p(G)\iff G'(p)=G'O^p(G)<G\iff G/G'(p)\ne 1.$$ The same applies to $H$.
Addendum: Since $G/G'(p)$ is abelian by definition, we have $G'\le G'(p)$. Since $G/G'(p)$ is a $p$-group, we also have $O^p(G)\le G'(p)$. Hence, $G'O^p(G)\le G'(p)$. Conversely, $G/G'O^p(G)$ is an abelian $p$-group and therefore $G'(p)\le G'O^p(G)$.