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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.2011-09-15
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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).2011-09-15
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    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).2011-09-15
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    @AmiteshDatta I have tried doing the exercises in Isaacs' Algebra, but I find progress to be very slow....2011-09-15
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    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.2011-09-15
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    @AmiteshDatta Too bad in our algebra class a lot of material is squeezed into so little time....I don't think we do even Sylow Theory.2011-09-15
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    @DBLim Indeed, this is unfortunate. 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, solvable and nilpotent groups, and transfer is all worth learning!2011-09-15

1 Answers 1

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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*?2011-09-15
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    Not the quotient. The set difference.2011-09-15
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    The solution is embarrassing simple.2011-09-15
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    @jspecter Sorry my bad got mixed up between forward and back slash. Been thinking too much about quotients!2011-09-15
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    @jug solution to what?2011-09-15
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    @D B Lim: Your problem (reducing to the cyclic case).2011-09-15
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    @D B Lim: (his comment got deleted and replaced during I posted my comment) Every proper subgroup $X$ of $G$ is contained in a maximal subgroup. $H$ is the only maximal subgroup, so $X \le H$. Now take $X = \langle x \rangle$.2011-09-15
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    @jspecter If we consider the cyclic subgroup generated by $\langle x \rangle < G$, then how can we just say from here that there must exist another maximal subgroup $K$ such that $H \subset K$?2011-09-15
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    @DBLim *Hint*: Note that every proper subgroup of a finite group is contained in a (proper) maximal subgroup. If $H$ is the (unique) maximal subgroup of $G$ and if $x\not\in H$, then we know that $\langle x \rangle$ is contained in a (proper) maximal subgroup of $G$.2011-09-15
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    @AmiteshDatta Well if the first bit about what you said is true we immediately have a contradiction that there was only one maximal subgroup in the beginning. Just some doubts concerning the first part of what you said, and about how $x$ must generate the whole group.2011-09-15
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    @DBLim The word "proper" is important. I should emphasize: every *proper* subgroup of $G$ is contained in a (proper) maximal subgroup of $G$.2011-09-15
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    @DBLim In particular, if a subgroup of $G$ is *not* contained in a (proper) maximal subgroup of $G$, then ...2011-09-15
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    @AmiteshDatta What do you mean by (proper) maximal subgroup?2011-09-15
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    @DBLim I think some students make the mistake of thinking that the group $G$ itself is the one and only example of a "maximal subgroup". I wished to stress (in case it was not clear) that $G$ is not considered to be an example of a "maximal subgroup of $G$"; indeed a subgroup $M$ of $G$ is maximal if and only if $M$ is a *proper subgroup* of $G$ and there is no *proper subgroup* of $G$ strictly containing $M$.2011-09-15
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    @DBLim Do you know how to answer your question now?2011-09-15
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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.2011-09-15
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    @AmiteshDatta I think I get it: A proper subgroup $H$ of $G$ is either maximal or is not. If it is not maximal, then by the negation of your definition we have that $H$ is contained strictly in some other maximal subgroup of $G$. Now back to the original problem, suppose that $x \in G\backslash H$, and consider $\langle x \rangle$. Assume that $\langle x \rangle < G$. Then $\langle x \rangle$ is a proper subgroup, and it cannot be maximal otherwise this would contradict our assumption. This means that $\langle x \rangle$ must be strictly contained in $H$. But then $x \notin H$ by assumption,2011-09-15
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    so either way we have a contradiction. In other words, $\langle x \rangle$ cannot be a proper subgroup of $G$, so it must be all of $G$.2011-09-15
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    @DBLim Yes! ${}$2011-09-15
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    @ArturoMagidin Now that was a result I did not know, I suppose it's a variation of something I do know: If a group $G$ has only one proper subgroup, then it must be normal.2011-09-15
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    @DBLim I would now advise you to do the exercise in Isaacs following this one (on the Frattini subgroup) as it will further your understanding of this technique.2011-09-15
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    @DBLim ***Exercise***: Let $G$ be a finite group with $\left|G\right|=n$. Let us assume that there is exactly one subgroup of $\left|G\right|$ of order $m$ for some positive integer factor $m$ of $n$. Prove that the subgroup of $G$ of order $m$ is a normal subgroup of $G$.2011-09-15
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    @ArturoMagidin Can you expand further to your comment above? Why must $\phi(H)$ be a maximal subgroup of $\phi(G)$?2011-09-15
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    @DBLim *Hint*: Note that $K$ is a proper subgroup of $G$ containing $\phi(H)$ if and only if $\phi^{-1}(K)$ is a proper subgroup of $G$ containing $H$ (since $\phi$ is an automorphism).2011-09-15
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    @AmiteshDatta If $H$ is a subgroup of $G$, then when we do $gHg^{-1}$ we get another subgroup of the same order. Since there is only one subgroup of this order, $gHg^{-1} = H$ and so $H$ is normal.2011-10-01
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    @DBLim Yes, you are correct.2011-10-02