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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.

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@GeoffRobinson At this point in the book Sylow Theory has not been introduced, so cannot be used. Furthermore I don't know about Sylow Theory yet. –  user38268 Sep 15 '11 at 12:46
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Of course, Sylow theory allows you to directly solve the problem rather than first reduce it to the case of cyclic groups (as in jspecter's answer). However, Sylow theory is powerful and it is probably a good idea to use it sparingly in simple problems such as this one (for this minimal use of Sylow theory trains one to better understand the simpler techniques of finite group theory). –  Amitesh Datta Sep 15 '11 at 13:22
    
Also, this exercise is in the set of chapter 2 exercises in Isaacs; I thought Exercise 2 in this set was a very good example of the importance of "subgroups" in finite group theory (and it is also an excellent exercise; Isaacs really must be commended for his choice of exercises). –  Amitesh Datta Sep 15 '11 at 13:32
    
@AmiteshDatta I have tried doing the exercises in Isaacs' Algebra, but I find progress to be very slow.... –  user38268 Sep 15 '11 at 13:36
    
Yes, the exercises in Isaacs are very good because they teach you technique in finite group theory. In particular, they are challenging because in many cases an exercise in Isaacs requires a technique not discussed in the text. However, the process of solving all the exercises in Isaacs is very rewarding and one that I would highly recommend. In my opinion, there are few better ways to learn the elementary techniques of finite group theory than to read a textbook of Isaacs and do most or all of the exercises. –  Amitesh Datta Sep 15 '11 at 13:46

1 Answer 1

up vote 5 down vote accepted

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 $G.$

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To even form the quotient, don't I need that $H$ be normal which I don't even know a priori? –  user38268 Sep 15 '11 at 12:48
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Not the quotient. The set difference. –  jspecter Sep 15 '11 at 12:48
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The solution is embarrassing simple. –  j.p. Sep 15 '11 at 12:53
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@jug solution to what? –  user38268 Sep 15 '11 at 12:56
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@DBLim: It might also be worth noting that if $H$ is the unique maximal subgroup of $G$, then it must be normal: if $\phi$ is any automorphism of $G$, then $\phi(H)$ is a maximal subgroup of $\phi(G)=G$, hence $\phi(H)=H$. Thus, $H$ is characteristic, and in particular normal. –  Arturo Magidin Sep 15 '11 at 13:43

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