# Existence of a non-abelian group of order $p^n$.

Question:

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

Attempt:

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?

• If you construct a nonabelian group of order p^3, the you can just do a direct product with a cyclic group of order $p^{n-3}$. – Mariano Suárez-Álvarez May 29 '15 at 23:26
• @Mariano sure, post a comment that's simpler than my answer why don't you! :-p – Matt Samuel May 29 '15 at 23:30
• @MarianoSuárez-Alvarez So you mean as $$((\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p )\times \mathbb Z_{p^{n-3}}$$ – Aaron Maroja May 29 '15 at 23:41
• 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. – Mariano Suárez-Álvarez May 30 '15 at 12:44
• @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. – 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.
• 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 $$((\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p )\times \mathbb Z_p$$ – Aaron Maroja May 30 '15 at 2:17
• 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 $n-2$! – mich95 May 30 '15 at 2:21