Subgroups of cyclic group of order $p^n$

Question

If $H$ and $K$ are subgroups of a cyclic group of order $p^n$ , where $p$ is prime , and $|H|>|K|$ , then is it true that $K \subset H $ ?

Answer 1

Hint: in a cyclic group of order $n$ for each divisor $d$ of $n$ there is a unique subgroup of order $d$.

Answer 2

In general, the subgroups of any cyclic group $G = \langle g \rangle$ of size $n$ are $H_d = \langle g^d \rangle$ of size $\frac{n}{d}$ for each $d$ dividing $n$. In particular they are all cyclic, and there is a unique subgroup corresponding to each divisor of $n$. See if you can prove this statement, it will make understanding (finite) cyclic groups much easier.

For $n = p^k$, say, the subgroups are thus $H_{p^j}=\langle g^{p^j}\rangle$ for each $0 \leq j \leq k$. As such, each one contains all smaller subgroups.

Answer 3

In a cyclic group of order $n$, for every $d \mid n$ there is an unique subgroup of order $d$, which is itself cyclic.

If $H,K$ are subgroups of a cyclic group $G$ of order $p^n$ with $|H| > |K| $, then $|K| \mid |H|$ and so $H$ contains a subgroup of cardinality $|K|$. But this subgroup is $|K|$ due to the uniqueness of subgroups in $G$ and so $$K \subset H$$