$\pi$-radical of group

Question

Let $G$ be abelian and periodic. Let $\mathbb{P}$ be the set of prime numbers, $\pi \subseteq \Bbb P$ and $\pi ^{\prime }=\Bbb P\setminus\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 if $G/O_{2^{\prime}}\left( G\right) $ is a $2$-group then $G$ is $2$-group?

Answer 1

In an abelian group, every subgroup is normal. (This is easy to see: if $ab=ba$ for all $a,b\in G$, then surely $aha^{-1}=h\in H$ for all $a\in G,h\in H$.) Thus no matter which set of primes you pick for $\pi$, $G/\mathcal{O}_{\pi^\prime}(G)$ must be a $\pi$-group. You can't conclude anything about $G$ from this.

$\mathcal{O}_\pi(G)$ and $\mathcal{O}_{\pi'}(G)$ are concepts invented to deal with nonabelian groups. They're only interesting when there are $\pi$-subgroups that aren't normal. Let's look at an example.

Consider the dihedral group of order $20$, $$D_{20}=\langle r,s|r^{10},s^2,s^{-1}rs=r^{-1}\rangle.$$ $D_{20}$ contains a unique normal subgroup of order $2$, $Z=\langle r^5 \rangle$. No subgroups of order $4$ are normal. In particular, there are five subgroups of order $4$: $H_i=Z\times \langle r^i s \rangle$ for $i=0,\ldots,4$. It's easy to see that each of these subgroups intersect exactly at $Z$.

So in this case, we see that $\mathcal{O}_{\{2\}}(D_{20})=Z$. If we take $D_{20}/\mathcal{O}_{\{2\}}(D_{20})$, we get $D_{10}$, and if we mod out by $\mathcal{O}_{\{2\}'}(D_{20})=\mathcal{O}_{\{5\}}(D_{20})$, we get the Klein $V$ group. This helps us understand the structure of $D_{20}$ a little bit- we see that $D_{20}\cong \mathbb{Z}_2 \times D_{10}$.

This example suggests a different definition for $\mathcal{O}_\pi(G)$. With just one prime $p$, $\mathcal{O}_{\{p\}}(G)$ is the intersection of all Sylow $p$-subgroups. Naturally, when Hall $\pi$-subgroups exist, $\mathcal{O}_{\pi}(G)$ is the intersection of those. (A caveat: Hall subgroups don't always exist for every subset of prime divisors when $G$ isn't solvable, in which case we must use the standard definition.) So the purpose of $\mathcal{O}_\pi$ is to teach us how the biggest $\pi$-subgroups overlap, and similarly for $\mathcal{O}_{\pi^\prime}$. You can think of it as the "invariant $\pi$-part of $G$."

So now that you've seen how the $\mathcal{O}_\pi$ and $\mathcal{O}_{\pi^\prime}$ groups are used, let's revisit abelian groups. You can see how the question becomes trivial. If $G$ is abelian, then every subgroup is normal, so there is always a unique Hall $\pi$-subgroup $H_\pi$ of every subset of primes $\pi$. Thus $\mathcal{O}_\pi(G)$ is the intersection of, well, just one group, so $\mathcal{O}_\pi(G)=H_\pi$. It follows that $G/\mathcal{O}_\pi(G)$ is a $\pi'$ group, and in fact $G/\mathcal{O}_\pi(G)\cong \mathcal{O}_{\pi^\prime}(G)$! The same goes for $G/\mathcal{O}_{\pi^\prime}(G)\cong \mathcal{O}_{\pi}(G)$. In other words, the $\pi$ and $\pi^\prime$ parts of an abelian group are completely separable from one another, no matter what which $\pi$ you choose. So $\mathcal{O}_\pi$ and $\mathcal{O}_{\pi^\prime}$ are not very interesting for abelian groups.

Answer 2

$G/O_{2'}(G)$ will always be a $2$-group.

In fact, if $a \in G$ has order $2^e t$, with $t$ odd, then $a^{2^e}$ has order $t$, so it is in $O_{2'}(G)$. Therefore $a O_{2'}(G)$ has order $2^e$ in $G/O_{2'}(G)$.

So for instance $G = \mathbf{Z}_6$ is not a $2$-group, but $G/O_{2'}(G) \cong \mathbf{Z}_2$ is a nontrivial $2$-group.

Note that $G$ being abelian, you do not need the normality assumption, and that you can rewrite $$ O_{\pi}(G) = \{ x \in G : \text{$x$ has an order which is a $\pi$-number} \}. $$