I am trying to use this fact in another proof, and I would like to make sure what I'm doing is correct first. Basically, I have a group, $G$ of order $p^k$, and I would like to show that there exist subgroups with order $p^i$ for all $i \leq k$. Here is what I have so far.
By Cauchy's Theorem, since $p^i$ divides $p^k$ for all $i \in \{1,2,…,k\}$, this means there is an element of order $p^i$. Thus, I can make a subgroup generated by this element, and it will also have order $p^i$ (which is a simple proof).
My issue with this is that it implies that $G$ is cyclic since there is an element of order $p^k$ by Cauchy. I have tried looking things up in several ways to try to disprove that, but I haven't found any good examples by googling. By the definition of a cyclic group, this would mean that $G$ is abelian too, and I am pretty sure that is not always the case? I think my understanding of Cauchy's theorem and cyclic groups in general must be wrong, but I'm not sure where I'm getting lost. I know that all groups of prime order are cyclic, but is that the case with groups of order of $p$ to some power?
Any tips/hints would be greatly appreciated! Thanks.
Best Answer
Use the class equation to show $G$ has a non-trivial center $Z$. Raise a non-identity element of $Z$ to a power such that the result $g\in Z$ has order $p$. Then $G/<g>$ is a smaller p-group and has subgroups of all smaller powers of $p$ by induction. Their pre-images provide the needed subgroups of $G$.
Note that the same proof works with "subgroup" replaced by "normal subgroup" throughout. Thus $G$ also has a normal subgroup of order $p^i$.