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

1. Jun 18, 2012

### tonit

1. The problem statement, all variables and given/known data

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 $K⊆H$.

3. The attempt at a solution

1 => 2.

Let $H,K$ be subgroups of $G = <g>$ where $o(g) = p^n$. We have $H = <g^a>$ and $K = <g^b>$ 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

2. Jun 18, 2012

### 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".

3. Jun 18, 2012

### 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?