# how to show that a group of order $p^k$ has subgroups of order $p^i$

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.

• Not all $p$-groups are cyclic. Indeed not all $p$-groups are Abelian. – Lord Shark the Unknown Oct 24 '18 at 4:40
• You might want to consider the group $\Bbb Z/2\Bbb Z \times \Bbb Z/\Bbb Z$. which has order $2^2$, but no element of order $2^2$; that suggests you're misapplying Cauchy's theorem. – John Hughes Oct 24 '18 at 4:41
• Cauchy's theorem only promises the existence of an element of order $p$ precisely. Higher order elements need not exist. – Jyrki Lahtonen Oct 24 '18 at 4:42
• To prove the existence of such a subgroup you apply Cauchy to the center of $G$ (known to be non-trivial). Then form the quotient (of order $p^{k-1}$) and apply a suitable the induction hypotesis. – Jyrki Lahtonen Oct 24 '18 at 4:44
• Ah okay! I tried induction before, but I didn't get the whole way through. Dumb question, but how did you know to apply it to the center of G? @JyrkiLahtonen – sadsloth_96 Oct 24 '18 at 4:47

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 $$G/$$ 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 $$p^i$$.