1
$\begingroup$

Let $K$ be a class of finite groups, then this class is called closed iff

i) homomorphic images of groups in $\mathcal K$ lie in $\mathcal K$,

ii) subgroups of groups from $\mathcal K$ lie in $\mathcal K$,

iii) direct products of groups form $\mathcal K$ lie in $\mathcal K$.

A finite group is called $\pi$-closed, if $G/O_{\pi}(G)$ is a $\pi'$-group, where $O_{\pi}(G)$ is the largest $\pi$-normal subgroup of $G$. For a set of primes $\pi$ a group is called a $\pi$-group if just primes from $\pi$ divide its order, and a $\pi'$-group if just primes not from $\pi$ divide its order. If we denote by $O^{\pi}(G)$ the smallest normal subgroup of $G$ such that its factor group is a $\pi$-group, then a group $G$ is $\pi$-closed precisely when $$ O_{\pi}(G) = O^{\pi'}(G). $$ I want to show that the class of all $\pi$-closed groups is a closed class (I guess this to be true).

I have found a proof for homomorphic images, but I am stuck on showing it for subgroups and direct products. If $N \unlhd G$, then $$ O_{\pi}(N) = N \cap O_{\pi}(G) $$ and so in this case a little rearranging with the isomorphism theorems yields that $N$ is $\pi$-closed. But I have no idea how to proceed for arbitrary subgroups of $G$, so any ideas? And also how to show it for direct products?

$\endgroup$
  • 2
    $\begingroup$ If $H \le G$ then clearly $H \cap O_\pi(G) \le O_\pi(H)$, and $H/(H \cap O_\pi(G)) \cong HO_\pi(G)/O_\pi(G)$ is a $\pi'$-group. It seems obvious that $O_\pi(G \times H) = O_\pi(G) \times O_\pi(H)$. $\endgroup$ – Derek Holt Apr 8 '15 at 13:55
  • 2
    $\begingroup$ A finite group is $\pi$-closed iff the set of all $\pi$-elements is a subgroup. $\endgroup$ – j.p. Apr 8 '15 at 14:02
  • $\begingroup$ @DerekHolt: For me $O_{\pi}(G)\times O_{\pi}(H) \le O_{\pi}(G\times H)$ is obvious, but $O_{\pi}(G\times H) \le O_{\pi}(H)\times O_{\pi}(G)$ not so? $\endgroup$ – StefanH Apr 8 '15 at 19:26
  • $\begingroup$ @Stefan: What is the relation between $O_\pi(G)$ and $O_\pi(G\times H)\cap G$ (considering $G$ as subgroup of $G\times H$)? $\endgroup$ – j.p. Apr 9 '15 at 6:32
  • 1
    $\begingroup$ @j.p. Let $X := p_1(O_{\pi}(G\times H)), Y := p_2(O_{\pi}(G\times H))$, then as $X,Y$ are normal and $\pi$-groups, we have $X \le O_{\pi}(G), Y \le O_{\pi}(H)$ and therefore $X\times Y \le O_{\pi}(G)\times O_{\pi}(H)$. Also for arbitrary sets $Z \le A\times B$ we have $Z \le p_1(Z)\times p_2(Z)$, and therefore $O_{\pi}(G\times H) \le X\times Y \le O_{\pi}(G)\times O_{\pi}(H)$, so this gives the other direction. Thanks! $\endgroup$ – StefanH Apr 9 '15 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.