7
$\begingroup$

I encountered the following exercise in Isaacs' Algebra:

"Suppose a group $G$ has only one maximal subgroup. Prove that the order of $G$ must be a power of a prime".

I think I've proven this for the case when $G$ is cyclic, based on the observation that in a cyclic group $G$ with a subgroup $H$, $H$ is maximal iff $\frac{|G|}{|H|}$ is prime. However if $G$ is not cyclic, I cannot use the property that there always exists a subgroup of order some divisor of $|G|$. I have run out of ideas in solving this problem, how can I proceed from here? Please do not post complete solutions.

  • 0
    @DBLim Indeed, this is unfortun$a$te. However, you can learn Sylow theory in the summer if you wish. I think it is certainly worth learning for its elegance and power. In fact, the material in Isaacs on permutation groups, solva$b$le $a$nd nilpotent groups, $a$nd transfer is all worth learning!2011-09-15

1 Answers 1

5

All such groups are cyclic. To see this let $H$ be the unique maximal subgroup of $G$ and $x\in G\setminus H.$ Then the subgroup generated by $x$ is contained in no maximal subgroup and hence is equal to $G.$

  • 0
    @DBLim Yes, you are correct.2011-10-02