True of false: For every integer $\displaystyle n \geq 2,$ there exists a non-abelian group of order $\displaystyle n^3.$

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- Jul 2nd 2009, 11:13 AMNonCommAlgAlgebra, Problems For Fun (30)
__True of false__: For every integer $\displaystyle n \geq 2,$ there exists a non-abelian group of order $\displaystyle n^3.$ - Jul 2nd 2009, 02:58 PMBruno J.
If $\displaystyle n$ is even, $\displaystyle D_{n^3/2}$ is nonabelian of order $\displaystyle n^3$.

If $\displaystyle n$ is odd, perhaps we can decompose $\displaystyle n$ into a product of prime powers and show that for every odd prime $\displaystyle p$ there exists a nonabelian group of order $\displaystyle p^3$... (Worried) - Jul 2nd 2009, 11:17 PMSwlabr
It is sufficient to prove the result for $\displaystyle n=p$, a prime number, as then cross products give us the result.

So, a non-abelian group of order $\displaystyle p^3$? Does the group with presentation $\displaystyle G=<x,a|x^p=1, a^{p^2}=1, a^x=a^{1+p}>$ work? Clearly it is non-abelian (as if it was $\displaystyle a^x=a \neq a^{1+p}$), and it has order $\displaystyle p^3$ (as $\displaystyle |G| = o(g_1) * \ldots * o(g_i)$ with $\displaystyle G=<g_1, \ldots, g_m>$ a minimal generating set and $\displaystyle o(g_i) < \infty$).

(This is a specific case of the group $\displaystyle G=<x,a|x^p=1, a^{p^{n-1}}=1, a^x=a^{1+p^{n-2}}>$, $\displaystyle n \geq 3$, the only non-abelian p-group that has a cyclic maximal subgroup and is not of maximal class - see Robinson, A Course in the Theory of Groups, section 5.3.4) - Jul 3rd 2009, 05:27 PMNonCommAlg
$\displaystyle G=\left \{\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z} \right \}.$

- Jul 3rd 2009, 11:25 PMSwlabr