Proving Cyclic Group Order as Power of Prime

[terhorst]

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!

 

[Dick]

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?

 

[Kummer]

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.

 

[Kummer]

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.