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Let $G$ be a finite group different from a cyclic $p-$group and $\Phi(G)=M_i\cap M_j$, where $M_i$ and $M_j$ are two arbitrary distinct maximal subgroups of $G$ and $i, j \geq 1$. Is it possible to characterize such $G$? ($\Phi(G)$ denotes the Frattini subgroup of $G$, which is the intersection of all maximal subgroups of $G$).

Examples: (a) Cyclic groups of order $p^nq^m$ are examples of such groups, where $p, q$ are distinct primes. Also quaternion group of order 8 is another example.

(b) Alternating groups $A_n; n\geq 5$, have no such property because as well as we know, there are different non-trivial intersections of any two point stabilizers in $A_n$ and on the other hand $\Phi(A_n)=1$.

Remark: According to an exercise we know that if $p$ is a prime divisor of $|G|$, ${\rm Syl}_p(G)$ is the set of all Sylow $p-$subgroups of $G$ and $P\in {\rm Syl}_p(G)$ then $P\nleq \Phi(G)$. Combining this fact with the condition on $G$ implies that each Sylow subgroup of $G$ contained in exactly one maximal subgroup of $G$. May be this point can help.

Thank's a lot!

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So you're asking for groups with exactly 2 maximal subgroups? If so, $Q_8$ does not work, since it has 3 maximal subgroups (generated by $i$, $j$, and $k$). –  Nishant May 25 at 13:11
    
Are you saying that there exist two maximal subgroups whose intersection is the Frattini, or that any two different maximal subgroups intersect in the Frattini? –  Geoff Robinson May 25 at 13:33
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I think the OP means to characterize those groups of which the Frattini subgroup can be realized as the intersection of two different maximal subgroups. –  Nicky Hekster May 25 at 13:45
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Well I think he wants the intersection of any pair of distinct maximal subgroups to be equal to $\Phi(G)$. But why waste time speculating? –  Derek Holt May 25 at 13:52
    
Dear Prof. Robinson my mean is that any two different maximal subgroups intersect in the Frattini subgroup, as Prof. Holt mentioned in his comment. –  sebastian May 25 at 14:01
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up vote 4 down vote accepted

I interpret your question as saying that any two maximal subgroups of $G$ intersect in $\Phi(G).$ I only consider the case $\Phi(G) =1,$ which is no real loss of generality, because $G/\Phi(G)$ has the same property.

If $G$ is nilpotent, then $G$ is Abelian of square free exponent as $\Phi(G) = 1$. If $G$ is a $p$-group, then $G$ is either cyclic or elementary Abelian of order $p^{2}.$ If $G$ is nilpotent but not a $p$-group, and $\Phi(G) = 1,$ then $G$ must be cyclic of order $pq$ for distinct primes $p$ and $q.$

Suppose then that $\Phi(G) = 1,$ but that $G$ is not nilpotent. Then $G$ has a maximal subgroup $H$ with $H \not \lhd G.$ Then $H = N_{G}(H),$ and $H \cap H^{g} = 1$ for all $g \in G \backslash H.$ By Frobenius's theorem, there is $K \lhd G$ with $G = KH$ and $K \cap H = 1.$ Also ${\rm gcd}(|H|,|K|) = 1.$ Since $H$ is a maximal subgroup of $G,$ no proper non-trivial subgroup of $K$ is $H$-invariant. Hence $K$ is an elementary Abelian $q$-group for some prime $q.$ Now $H$ must have a unique maximal subgroup, for if $S$ and $T$ are different maximal subgroups of $H,$ then $KS$ and $KT$ are distinct maximal subgroups of $G$ with intersection $K \neq 1.$ Hence $H$ is a cyclic $p$-group for some prime $q \neq p.$ In particular, $G$ is solvable. (Later edit: In fact $H$ has order $p$: I thought I had an argument for this earlier, but realised it was faulty: I think this one is OK. The hypotheses imply that every elemnt of order $p$ is contained in a unique maximal subgroup of $G$: if $|H| >p,$ then $ \Omega_{1}(H) (\neq 1,H) $ is contained in the maximal subgroup $H$, and in a maximal subgroup containing $K \Omega_{1}(H),$ and these are different).

Back to the general case (maybe $\Phi(G) \neq 1),$ we may conclude that $G$ is solvable, since $\Phi(G)$ is nilpotent. Then $|G|$ must have the form $p^{a}q^{b}$ with $p,q$ distinct primes ( but possibly $ab = 0),$ for if $G$ were divisible by $3$ different primes $p,q$ and $r,$ then a Sylow $p$-subgroup of $G$ would be contained in a Hall $q^{\prime}$-subgroup and a Hall $r^{\prime}$-subgroup, and no maximal subgroup of $G$ contains both of these.

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a suggestion, if you divides your answer into the parts, it becomes easy to read. –  mesel May 25 at 14:06
    
Yes, thanks, I tried to do this, but at first the latex did not respect my paragraph breaks. –  Geoff Robinson May 25 at 14:16
    
Dear prof. Robinson, thank you very much for your nice and comprehensive answer! –  sebastian May 25 at 22:16
    
Dear Prof. Robinson, excuse me! I tried to show that if $S$ and $T$ are maximal subgroups of $H$, then $KS$ and $KT$ are distinct maximal subgroups of $G$, but I couldn't. Is it possible to explain it more? Thank you very much! –  sebastian May 27 at 20:45
    
Well, they have different images in $G/K,$ so they must be different subgroups of $G.$ –  Geoff Robinson May 27 at 20:48
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