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?

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Your choice of notation makes your question very difficult to read. Usually $\pi^{\prime}$ would denote the complementary set of primes to $\pi,$ and also $O^{\pi}(G)$ would denote the smallest norrmal subgroup of $G$ for which the factor group is a $\pi$-group. – Geoff Robinson Aug 30 '11 at 14:12
@Geoff Robinson: you're right, I forgot $\pi'$ denotes the complementary set of primes, sorry. I've modified it now. – user6495 Aug 30 '11 at 14:16
@user6495: are you defining π-solvable to mean a solvable group that is also a π-group? If so, it is not true. If you mean a π-solvable group is a group whose chief factors are either p-groups for p in π or π′-groups, then it is true, simply because P-solvable groups are also π-solvable. – Jack Schmidt Aug 30 '11 at 14:45
@Jack Schmidt: I mean that $G/O^{\pi}(G)$ is solvable and also a $\pi$-group. Is it possible to reduce the composition $O^{\pi}(O^{p}(G))$? – user6495 Aug 30 '11 at 14:47

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.

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Hopefully I have not introduced any errors... – Zev Chonoles Aug 30 '11 at 16:01