# Exactly one nontrivial proper subgroup

Question: Determine all the finite groups that have exactly one nontrivial proper subgroup.

MY attempt is that the order of group G has to be a positive nonprime integer n which has only one divisor since any divisor a of n will form a proper subgroup of order a. Since 4 is the only nonprime number that has only 1 divisor which is 2, All groups of order 4 has only 1 nontrivial proper subgroups (Z4 and D4)

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$D_4$ is the Klein $4$-group, which has three non-trivial proper subgroups. – Brian M. Scott Sep 18 '12 at 21:34
What about $9$? – Mariano Suárez-Alvarez Sep 18 '12 at 21:36
Or 25?    – Did Sep 18 '12 at 21:38
Your claim that any divisor of $n$ gives you a subgroup of that order is false: for example, $A_4$ has order $12$ but no subgroup of order $6$. – Chris Eagle Sep 18 '12 at 21:40
Hint: Pick an element outside your unique proper subgroup, and check the subgroup it generates... – user641 Sep 19 '12 at 4:29

Let $H$ be the only non-trivial proper subgroup of the finite group $G$. Since $H$ is proper, there must exist an $x \notin H$. Now consider the subgroup $\langle x\rangle$ of $G$. This subgroup cannot be equal to $H$, nor is it trivial, hence $\langle x\rangle = G$, that is $G$ is cyclic, say of order $n$. The number of subgroups of a cyclic group of order $n$ equals the number of divisors of $n$. So $n$ must have three divisors. This can only be the case if $n$ is the square of a prime number. So, $G \cong C_{p^2}$.

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You began reasonably, but then you went off the track. First, $D_4$ won’t work: it’s the Klein $4$-group, $(\Bbb Z/2\Bbb Z)\times(\Bbb Z/2\Bbb Z)$, which has three non-trivial proper subgroups, one generated by each of the non-identity elements.

Secondly, the fact that some integer $n$ divides the order of $G$ does not ensure that $G$ has a subgroup of order $d$, as noted in the comments. You do, however, have the first Sylow theorem available. (Note: This replaces the nonsense that I wrote originally.)

Finally, $4$ isn’t the only non-prime with only one non-trivial divisor: for each prime $p$, $p^2$ is such a number. You want at least the groups $\Bbb Z/p^2\Bbb Z$, or in your notation $\Bbb Z_{p^2}$, for all primes $p$. Can you identify exactly what the non-trivial proper subgroup is in each of these groups?

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@anon: Yes, I realized almost immediately that I’d written an idiocy. You apparently caught it just about when I did. – Brian M. Scott Sep 18 '12 at 21:53

Hints/questions to consider: Have you done Sylow's theorem? How many Sylow subgroups can your group have? Show that your group must be cyclic- (hint for this: if not, all its element are in the unique proper subgroup). How many subgroups does a cyclic group of order $p^{n}$ have, when $p$ is prime and $n$ is a positive integer?

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Why do we need Sylow's to think about cyclic groups? – anon Sep 18 '12 at 22:07
We don't, but a priori such a finite group could be anything. We use Sylow's theorem to exclude the non-cyclic groups from consideration. – kahen Sep 18 '12 at 22:11
@anon: it's a question of in what sequence you do things: you could indeed first prove the group cyclic, then conclude it must have prime power order without explicitly making use of Sylow's theorem. – Geoff Robinson Sep 18 '12 at 22:14
@kahen As Geoff hints at, a very simple argument shows the group must be cyclic - by only considering cyclic subgroups. I feel Sylow is too high-powered for this exercise; in fact I suspect this is an exercise from Contemporary Abstract Algebra's chapter on cyclic groups, which is way before Sylow tech is even touchable. [Edit: Yes, it is exercise 34 in ch 1-4 supplementary exercises.] – anon Sep 18 '12 at 22:17
@anon: I think it is probably the case that almost anyone who HAD done Sylow would think of using it. I asked Johnny whether he had done Sylow's theorem, because the problem might be easier with Sylow than without, but it certainly can be done without. – Geoff Robinson Sep 18 '12 at 22:40