Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Hopefully I have not introduced any errors... –  Zev Chonoles Aug 30 '11 at 16:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.