Algebra, Problems For Fun (30)

NonCommAlg · #1 $Old$ July 2nd, 2009, 01:13 PM

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

Bruno J. · #2 $Old$ July 2nd, 2009, 04:58 PM

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 $p^3$ ...

Swlabr · #3 $Old$ July 3rd, 2009, 01:17 AM

Quote:

Originally Posted by Bruno J. $View Post$

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 $p^3$ ...

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 $G=<x,a|x^p=1, a^{p^2}=1, a^x=a^{1+p}>$ 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 $G=<g_1, \ldots, g_m>$ a minimal generating set and $o(g_i) < \infty$ ).

(This is a specific case of the group $G=<x,a|x^p=1, a^{p^{n-1}}=1, a^x=a^{1+p^{n-2}}>$ , $n \geq 3$ , 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)

NonCommAlg · #4 $Old$ July 3rd, 2009, 07:27 PM

Swlabr · #5 $Old$ July 4th, 2009, 01:25 AM

Quote:

Originally Posted by NonCommAlg $View Post$

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

That group is slightly more intuitively obvious...