I am reading I. Martin Isaacs' beautiful book Finite Group Theory. On p. 287 is a lemma that includes a condition (that $P$ is normal; notation fixed below) that I cannot find a need for in the proof.
My question: Am I missing something? How does the proof require $P$ to be normal? Or else, is this condition superfluous?
Fixing notation: given a finite group $G$ and a prime $p$, let $\mathbf{O}^p(G)$ be the unique minimal subgroup of $G$ such that the quotient is a $p$-group. If $S$ is a subgroup, the relation $S\triangleleft G$ means $S$ is normal in $G$, while $S\triangleleft\triangleleft G$ means $S$ is subnormal in $G$, i.e. there is an ascending sequence of subgroups $S=S_1< S_2< \dots< S_r = G$ where each $S_i$ is normal in $S_{i+1}$. The symbol $<$ means proper containment between subgroups.
Here is the statement and proof of the lemma from p. 287 in Isaacs' book.
9.26. Lemma. Let $G = SP$, where $G$ is a finite group, and where $S\triangleleft\triangleleft G$ and $P\triangleleft G$, and $P$ is a $p$-group for some prime $p$. Then $\mathbf{O}^p(G) = \mathbf{O}^p(S)$.
Proof. If $S=G$, there is nothing to prove, so we can assume $S<G$, and we choose $M$ with $S\subseteq M\triangleleft G$ and $M<G$. Then $M=S(M\cap P)$, and so working by induction on $|G|$ and applying the inductive hypothesis to $M$, we have $\mathbf{O}^p(M) = \mathbf{O}^p(S)$, and it suffices to show that $\mathbf{O}^p(M) = \mathbf{O}^p(G)$.
Now $\mathbf{O}^p(M)\triangleleft G$, and we write $\overline G = G/\mathbf{O}^p(M)$. Since $S\subseteq M$, we have $G=MP$, and thus $\overline G = \overline M\, \overline P$, which is a $p$-group. It follows that $\mathbf{O}^p(G)\subseteq \mathbf{O}^p(M)$. The reverse containment holds because $M/\mathbf{O}^p(G)$ is a $p$-group, and this completes the proof.
I can't see any need for the hypothesis that $P$ is normal in this proof! The hypothesis that $S$ is subnormal is needed in order for $M$ to exist, and the weight of the first paragraph is reduction to the case that $S$ (now replaced with $M$) is normal. This is already enough to know $\mathbf{O}^p(M)$ is normal in $G$ (as it is a characteristic subgroup of a normal subgroup). Then $\overline G = G/\mathbf{O}^p(M)$ exists. Then $\overline M$, which is $M$'s image under this canonical map, is a $p$-group, by definition of $\mathbf{O}^p(M)$. Meanwhile, $\overline P$, the image of $P$, is obviously also a $p$-group, and the cardinality of $\overline G = \overline M \,\overline P$ is thus a factor of a product of powers of $p$ and thus a power of $p$. Why isn't this all that's needed?
Some context: This proof follows the pattern of an earlier result, Lemma 9.15 found on p. 281 with the proof on pp. 281-2. It is the exact same statement except only assuming $P$ is nilpotent (rather than a $p$-group), and therefore with $S^\infty, G^\infty$ in place of $\mathbf{O}^p(S), \mathbf{O}^p(G)$. ($S^\infty, G^\infty$ are the unique minimal subgroups such that the quotients $S/S^\infty$ and $G/G^\infty$ are nilpotent.) The proof is also essentially the same, but in the earlier case, I do see the way the proof uses the normality of $P$: in order to conclude $\overline G = \overline M\, \overline P$ is nilpotent from the nilpotence of $\overline M, \overline P$, Isaacs argues that since they are both normal, they are both contained in the Fitting subgroup of $\overline G$, and thus so is their product. But I don't see a similar need in the $p$-group situation -- what am I missing?