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Let $\mathbb{P}$ is a set primes numbers, $\pi \subseteq P$ and $\pi ^{\prime }=P-\pi $

Let $O_{\pi }\left( G\right) =\left\langle N~;~N\trianglelefteq G\text{ and }% N\text{ is }\pi \text{-subgroup}\right\rangle $

Is it true that $O_{\pi }\left( G\right) =1\Longrightarrow G$ is $\pi ^{\prime }-$group?

if it is true, could not it have other $H$ $\pi $-subgroup (not normal) in $G$?

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1 Answer 1

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No. Let $\pi=\{2\}$ and $G$ be non-abelian of order 6. Then $O_\pi(G) = 1$, but $|G|$ is divisible by 2.

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Perhaps you can find a counterexample. Using the same $G$. –  Jack Schmidt Aug 15 '12 at 18:18
    
And $O_{\pi ^{\prime }}\left( G\right) =1\Longrightarrow G$ is $% \pi -$group?Is it worth the same argument for that case? –  User2040 Aug 15 '12 at 18:20
    
Lima: Thank you! –  User2040 Aug 15 '12 at 18:22

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