Proving Cyclic Group Order as Power of Prime

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Problem
Suppose for all subgroups H,K of a finite group G, either H \subset K or K \subset H. Show that G is cyclic and its order is the power of a prime.

Attempt
I think I get the intuition: if H and K are not the same, then one of them must be the trivial subgroup and the other must be G itself. So if g \in G but g \notin H, then \left\langle g \right \rangle is a subgroup containing g, so by hypothesis, H \subset \left\langle g \right \rangle. From here I want to show that H is actually the trivial subgroup. No idea yet about the power of a prime thing. Can anyone provide a hint? Thanks!
 
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Your intuition is not quite correct. Suppose H is a proper subgroup of G. Consider the subgroups H_g=gHg^(-1) for g in G. What's the containment relation between H and H_g? What kind of a subgroup is H? This should be enough of a hint to get you started. For the prime thing, if p and q are prime factors of G, then we know there are subgroups Hp and Hq of order p and q respectively. What is their containment relation?
 
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There is a minor mistake. You should have said, G is a non-trivial group.
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My Own Question: If the same conditions apply for an INFINITE group G, does the it mean it is cyclic? (Possibly Zorn's Lemma).
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Here is the answer to your question.
Say p\not = q are two primes dividing G. By Cauchy's theorem there are subgroups H \mbox{ and }K respectively so that |H|=p \mbox{ and }|K|=q. But how can (without lose of generality) H\subseteq K since all its non-identity elements are of order p while all the non-identity elements of K are of order q. A contradiction. Hence two distint primes do not divide G. And so |G|=p^n. Now we show that G\simeq \mathbb{Z}_{p^n}. Assume that G is not cyclic. Let a\not = e be any element, then \left< a \right> does not exhaust G by assumption. So choose b\in G \mbox{ with }b\not \in \left< a \right> and construct \left< b \right>. But hypothesis \left< a \right> \subseteq \left< b\right> \mbox{ or }\left< b \right> \subseteq \left< a\right>. But that is a contradiction. Q.E.D.
 
I made a mistake in my other post when I said \left< a \right> \subseteq \left< b\right> \mbox{ or }\left< b \right> \subseteq \left< a\right> is a contradiction. It is not. It however means that \left< a\right> \subseteq \left< b \right> since b\not \in \left< a \right> \implies \left< b\right> \not \subseteq \left< a\right>. Thus, \left< a\right> \subset \left< b \right>. So we can choose c\in G \not \in \left< b \right> to get \left< a\right> \subset \left< b \right> \subset \left< c\right>. Thus, continuing this we can an ascending chain condition of subgroups properly contained in another. Since G is finite it means this chain must terminate and hence there is an element which generates the full group.
Q.E.D.
 
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