# Pi-subgroup part-2

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

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

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

I think so, because every subgroup is normal in $G$.

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Yes, this is correct assuming $G$ is finite (or at least torsion), for exactly the reason you say (and Sylow's theorem). – Jack Schmidt Aug 15 '12 at 19:39
Lima: Thank you! I am needing that result. – User2040 Aug 15 '12 at 20:09
You can relax your condition of G being abelian to G being $\pi$-separable (that is G has a normal series with each factor either a $\pi$- or $\pi$'-group. – Nicky Hekster Aug 15 '12 at 21:57