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.

I have heard that the following is a conjecture due to Thompson: The number of maximal subgroups of a (finite) group $G$ does not exceed the order $|G|$ of the group.

My question is: did Thompson really conjecture this? If so, is there any literature on the subject?

share|improve this question

1 Answer 1

up vote 7 down vote accepted

I've seen this conjecture attributed to Wall (1961), for example: On a conjecture of G.E. Wall. This is a recent article (journal version appeared in 2007), and it gives a bunch of references. The conjecture remains open. Here is a very recent article which does not attack the conjecture itself, but uses it as an inspiration for a different conjecture.

share|improve this answer
5  
Wall proved it for solvable groups. Another proof (for solvable groups) is given in Pál Hegedűs, The number of maximal subgroups of a solvable group, Publ. Math. Debrecen 78 (2011), no. 3-4, 687–689, MR2867210. Another paper on the solvable case is Benjamin Newton, On the number of maximal subgroups of a finite solvable group, Arch. Math. (Basel) 96 (2011), no. 6, 501–506, MR2821467 (2012f:20092). –  Gerry Myerson Jun 21 '12 at 0:35
    
Many thanks for the references. –  the_fox Jun 21 '12 at 1:02
    
Question: if $P$ is a $p$-group and $M_1,M_2$ are maximal subgroups of $P$ such that $M_1/\Phi(P )=M_2/\Phi(P )$, does it follow that $M_1=M_2$? I need this in order to understand something Newton mentions in the first page of his paper. –  the_fox Jun 21 '12 at 15:00
    
@Stefanos Yes, and this holds in greater generality: if $H$ is a normal subgroup of $G$, and $G_1,G_2$ are subgroups containing $H$, then $G_1/H=G_2/H$ if and only if $G_1=G_2$. Indeed, if there exists $a\in G_1\setminus G_2$, then the coset $aH$ is contained in $G_1/H$ but not in $G_2/H$. –  user31373 Jun 21 '12 at 15:23
    
Thanks! My question was a bit silly, I have to admit! –  the_fox Jun 21 '12 at 15:40

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.