G is cyclic and |G| = p^n, p is prime <=> H,K Subgroups, H⊆K or K⊆H

[tonit]

Homework Statement

Show that the following conditions are equivalent for a finite group G:

1.G is cyclic and |G| = p^n where p is prime and n\geq 0
2.If H and K are subgroups of G, either H⊆K or K⊆H.

The Attempt at a Solution

1 => 2.

Let H,K be subgroups of G = &lt;g&gt; where o(g) = p^n. We have H = &lt;g^a&gt; and K = &lt;g^b&gt; where a and b divide p^n. Since p is prime, a = p^s and b = p^t. If s \leq t, this means a|b whence H⊆K. Similarly, if b \leq a we have K⊆H.

Now I'm stuck at 2 => 1. Any help is appreciated

 

[algebrat]

Maybe if you can use a fact like, for any integer that divides the group G, there is a subgroup of that order? Do you have a fact like that that you are permitted to use? Maybe only for prime divisors. That result I think comes sooner.

In other words, a strategy that might have lead you to this is to think about the negation of 1, so what if the order of G is not a power of a prime. And how this might lead to subgroups which are not "one a subset of another".

 

[tonit]

If H and K are subgroups of G, suppose H⊆K. So we have that |H| divides |K|, and they both divide |G|. If it would happen that |G| = p^ka^r where p, a are different primes, then G would have two subgroups M and N such that |M| divides p, |N| divides a, M ⊄ N and N ⊄ M, thus contradicting the condition.
So |G| = p^k for a prime p and an integer k.

Is it correct?
Now how do I show that G is cyclic?