# Algebra, Problems For Fun (30)

• July 2nd 2009, 12:13 PM
NonCommAlg
Algebra, Problems For Fun (30)
True of false: For every integer $n \geq 2,$ there exists a non-abelian group of order $n^3.$
• July 2nd 2009, 03:58 PM
Bruno J.
If $n$ is even, $D_{n^3/2}$ is nonabelian of order $n^3$.

If $n$ is odd, perhaps we can decompose $n$ into a product of prime powers and show that for every odd prime $p$ there exists a nonabelian group of order $p^3$... (Worried)
• July 3rd 2009, 12:17 AM
Swlabr
Quote:

Originally Posted by Bruno J.
If $n$ is even, $D_{n^3/2}$ is nonabelian of order $n^3$.

If $n$ is odd, perhaps we can decompose $n$ into a product of prime powers and show that for every odd prime $p$ there exists a nonabelian group of order $p^3$... (Worried)

It is sufficient to prove the result for $n=p$, a prime number, as then cross products give us the result.

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

(This is a specific case of the group $G=$, $n \geq 3$, 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)
• July 3rd 2009, 06:27 PM
NonCommAlg
$G=\left \{\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z} \right \}.$
• July 4th 2009, 12:25 AM
Swlabr
Quote:

Originally Posted by NonCommAlg
$G=\left \{\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z} \right \}.$

That group is slightly more intuitively obvious...