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 $G$ 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?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
    
Lima:It was great your explanation! –  User2040 Mar 5 '13 at 14:45
add comment

$G/O_{2'}(G)$ 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 $$ O_{\pi}(G) = \{ x \in G : \text{$x$ has an order which is a $\pi$-number} \}. $$

share|improve this answer
    
Lima: thank you! –  User2040 Mar 3 '13 at 17:25
add comment

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.