# Minimal Subgroups

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?

• 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? Sep 12, 2016 at 17:14
• 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. Sep 12, 2016 at 18:56
• 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. Sep 13, 2016 at 9:57

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.