4
$\begingroup$

Let $G$ be a finite group with the following property:

For every nontrivial proper subgroups $H$ and $K$ for which $H\cap K=1$, if the number of their minimal subgroups are $m$ and $n$ respectively, then the number of minimal subgroups of the subgroup generated by $H$ and $K$,$\langle H,K\rangle$, is $m+n$.

What are the possible structures for $G$? I found only cyclic groups and direct product product of a generalized quaternion group and a cyclic group of some odd order. Is there any other case?

$\endgroup$
  • $\begingroup$ This is badly worded. It is not clear from the structure of the sentence what "where $H \cap K = 1$" means. Perhaps the sentence should start "For every nontrivial proper subgroups $H$ and $K$ for which $H \cap K = 1$ ..."? $\endgroup$ – Derek Holt Sep 12 '16 at 15:33
  • $\begingroup$ Yes you are right. I corrected it. $\endgroup$ – H.Shahsavari Sep 12 '16 at 15:38
  • $\begingroup$ I think in such groups, every abelian subgroup is cyclic. for this, if an abelian subgroup A is not cyclic, then it contains at least two isomorphic minimal subgroups say $L_{1},L_2$, such that $L_{i}\cong Z_p$ for $i=1,2$ and some prime $p$. Now $L_{1}\times L_{2}$ contains $p+1$ minimal subgroups isomorphic to $Z_p$, a contradiction. Does it help? $\endgroup$ – H.Shahsavari Sep 12 '16 at 17:14
  • 1
    $\begingroup$ There must be a unique subgroup of order $p$ for all primes $p$ dividing $|G|$. But there are some other examples. $\mathtt{SmallGroup}(72,3)$, which has a normal subgroup $Q_8$ and cyclic Sylow $3$-subgroup is one such. Its centre has order $6$ and it has only two minimal subgroups, or orders $2$ and $3$, respectively. $\endgroup$ – Derek Holt Sep 12 '16 at 18:56
  • $\begingroup$ By Derek's comment, we get that every minimal subgroup is characteristic. There is no classification of finite groups whose all minimal subgroups are characteristic, but I guess in a paper, finite groups whose minimal subgroups are all normal was classified. If we find it then we may get some helpful results about groups which I am studying. $\endgroup$ – H.Shahsavari Sep 13 '16 at 9:57
2
$\begingroup$

We show that $G$ satisfies the given properties iff $G$ has a unique subgroup of order $p$ for every prime divider $p$ of $|G|$ and that every two Sylow subgroups of distinct orders permute.

Necessity: If $A$ and $B$ are distinct cyclic subgroups of prime order $p$, then $[A,B]\neq1$ otherwise $\langle A,B\rangle\cong C_p\times C_p$ has $p+1>2$ minimal subgroups, which contradicts the assumption. But then $\langle A,B\rangle$ has at least $p+1$ minimal subgroups including the conjugates of $A$ by elements of $B$ and $B$ itself, which is another contradiction. This shows that $G$ has a unique subgroup of order $p$ for every prime $p$ dividing $|G|$. Now if $P$ and $Q$ are a Sylow $p$-subgroup and a Sylow $q$-subgroup of $G$, respectively, with $p\neq q$, then $P\cap Q=1$ and since by assumption $\langle P,Q\rangle$ is a $\{p,q\}$-group, we must have $\langle P,Q\rangle=PQ=QP$, as required.

Sufficiency: Let $H$ and $K$ be two subgroups satisfying $H\cap K=1$. Since $G$ has a unique subgroup of any prime order dividing $|G|$, we must have $\gcd(|H|,|K|)=1$. Then $H\subseteq\prod_{i\in\pi(H)}P_i$ and $K\subseteq\prod_{i\in\pi(K)}P_i$ for suitable Sylow subgroups $P_i$. Since $P_i$ and $P_j$ permute pairwise, we have $\langle H,K\rangle\subseteq\prod_{i\in\pi(H)\cup\pi(K)}P_i$, from which the result follows.

Note that by a theorem of Suzuki all these groups are solvable (See MR0074411).

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.