Question about solvable group Let $\pi$, $P$ sets of prime numbers such that $\pi \subseteq P$.
Definition: $O^{\pi}(G)$ is the smallest normal subgroup of $G$ such that $G/O^{\pi}(G)$ is a $\pi$-solvable group.
Question: Is it true that $O^{\pi}(G)=O^{\pi}(O^{P}(G))$?
We have the inclusion $O^{\pi}(O^{P}(G)) \subseteq O^{\pi}(G)$.
Note $G/O^{\pi}(O^{P}(G))$ is solvable so we are done if we can show that $G/O^{\pi}(O^{P}(G))$ is a $\pi$-group. But is this true?
 A: I'll give a counterexample to the original, a possible typo that is true, and then a direct answer to your simplification question.
Cex: Let $G$ be cyclic of order 6, let $P =\{ 2, 3 \}$, and $π = \{ 2 \}$.  Then $O^P( G) = 1$ is the identity subgroup, since $G$ itself is a solvable $P$-group.  Of course, $O^π(O^P( G ) ) = O^π( 1 ) = 1$ as well.  However, $O^π(G )$ is cyclic of order 3, so that the quotient is a $π$-group (a 2-group).
True: On the other hand, I believe $O^P(O^π( G ) ) = O^P(G )$ in general, since solvable $P$-groups is a class closed under extensions.  In more detail: $G/O^π(G)$ is a solvable $π$-group, so also a solvable $P$-group.  $O^π(G) / O^P(O^π(G) )$ is by definition a solvable $P$-group, and since an extension of a solvable $P$-group by a solvable $P$-group is a solvable $P$-group, one has $G/O^P(O^π(G))$ is a solvable $P$-group.  Hence $O^P(G) ≤ O^P(O^π(G))$. Since solvable $P$-groups are closed under normal subgroups, one has $O^P(O^π(G)) ≤  O^P(G)$.  Hence $O^P(O^π(G)) =  O^P(G)$.
Simplify: It might be that $O^π(O^P(G)) = O^P(G)$.  One has $≤$ by definition.  $O^P(G)/O^π(O^P(G))$ is a solvable $π$-group, so also a solvable $P$-group, so $G/O^π( O^P(G))$ is a solvable $P$-group, and $O^P(G) ≤ O^π(O^P(G))$.
Note that "true" and "simplify" look similar, but that the hypotheses on $π$ and $P$ are not symmetric.
These should apply to any normal-subgroup and extension closed formations.
