Let $p$ be any prime and $n \geq 3$. Show that there exists a non-abelian group of order $p^n$.


Take $n = 3$. Writing $\mathbb Z_p \times \mathbb Z_p = \{e, \alpha_1, \ldots, \alpha_{p^2 - 1}\}$ and considering $$\begin{align}\tau_{jk} : \mathbb Z_p \times \mathbb Z_p&\to \mathrm {Aut} (\mathbb Z_p)\,\\e &\mapsto id\\\alpha_j &\mapsto id \\\alpha_i &\mapsto \rho_k \end{align}$$

$\forall i \neq j$ where $\rho_k : \mathbb Z_p \to \mathbb Z_p$ is such that $\rho_k (1) = k$, with $k \in \{2, \ldots, p-1\}$. Then we take the group $(\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p$.

Then $$(\alpha_i , 2) \ltimes_{\tau_{jk}} (\alpha_i, 1) = (2\alpha_i, 1 + \tau_{jk} (\alpha_i)(2)) = (2\alpha_i , 1 +2 k)$$

and $$(\alpha_i , 1) \ltimes_{\tau_{jk}} (\alpha_i, 2) = (2\alpha_i, 1 + \tau_{jk} (\alpha_i)(2)) = (2\alpha_i , 1 +k)$$

taking $k = 1$ for example we have that the group $(\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p$ is not abelian.

  • I was wondering if this is a good aproach. If not, is there an easier way to reach the result?

  • If this is the "best" way then would the higher cases ($n > 3$) be more of a notation play game?

  • 3
    $\begingroup$ If you construct a nonabelian group of order p^3, the you can just do a direct product with a cyclic group of order $p^{n-3}$. $\endgroup$ – Mariano Suárez-Álvarez May 29 '15 at 23:26
  • 1
    $\begingroup$ @Mariano sure, post a comment that's simpler than my answer why don't you! :-p $\endgroup$ – Matt Samuel May 29 '15 at 23:30
  • 1
    $\begingroup$ @MarianoSuárez-Alvarez So you mean as $$((\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p )\times \mathbb Z_{p^{n-3}}$$ $\endgroup$ – Aaron Maroja May 29 '15 at 23:41
  • $\begingroup$ Your description of an example for n=3 does not look OK. The function $\tau$ that you tried to define is not well-defined (your list of the elements of the group has one element too many, for starters!) so you did not really construct a group at all. $\endgroup$ – Mariano Suárez-Álvarez May 30 '15 at 12:44
  • $\begingroup$ @MarianoSuárez-Alvarez Sorry, but I didn't follow the "list of the elements of the group has one element too many" part. Thanks for you help. $\endgroup$ – Aaron Maroja May 30 '15 at 12:52

Take a vector space over the field of order $p$ of dimension $n-1$ and take the semidirect product with $\mathbb Z_p$. If you can use the fact that the order of the general linear group is divisible by $p$, you can use Cauchy's theorem to see that we can map a generator to a matrix of multiplicative order $p$ without even having to explicitly construct one.


Let $G$ be the group of unipotent $3 \times 3$ matrices with entries in $\mathbf{F}_{p}$ (i.e. the upper triangular $3 \times 3$ matrices with $1$'s on the diagonal). It's a non abelian group of order $p^{3}$, as $p$ is odd. Consider $G \times \mathbf{Z}/p^{n-3}\mathbf{Z}$, it's a non abelian group.

  • $\begingroup$ So basically I could have done the same thing with $$(\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p$$ and then get $$((\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p )\times \mathbb Z_p$$ $\endgroup$ – Aaron Maroja May 30 '15 at 2:17
  • $\begingroup$ Jup, as long as you find a non abelian group of order $p^{3}$! I forgot to mention that if p=2, you can take $D_{8}$ with a cyclic 2 group of order $n-2$! $\endgroup$ – mich95 May 30 '15 at 2:21
  • $\begingroup$ Is my non-abelian group okay? $\endgroup$ – Aaron Maroja May 30 '15 at 2:22
  • $\begingroup$ I have reread the proof, and while rereading it, I found it nice, but did not check line by line. But I think, you're assuming that p is odd, or am I mistaken ? $\endgroup$ – mich95 May 30 '15 at 2:24
  • $\begingroup$ See also math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf. $\endgroup$ – lhf May 30 '15 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.