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.

  • $\begingroup$ Not all $p$-groups are cyclic. Indeed not all $p$-groups are Abelian. $\endgroup$ – Angina Seng Oct 24 '18 at 4:40
  • $\begingroup$ 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. $\endgroup$ – John Hughes Oct 24 '18 at 4:41
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    $\begingroup$ Cauchy's theorem only promises the existence of an element of order $p$ precisely. Higher order elements need not exist. $\endgroup$ – Jyrki Lahtonen Oct 24 '18 at 4:42
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 24 '18 at 4:44
  • $\begingroup$ 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 $\endgroup$ – 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 $G/<g>$ 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 $p^i$.


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