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Thread: Algebra, Problems For Fun (30)

  1. #1
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    Algebra, Problems For Fun (30)

    True of false: For every integer $\displaystyle n \geq 2,$ there exists a non-abelian group of order $\displaystyle n^3.$
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    MHF Contributor Bruno J.'s Avatar
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    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 $\displaystyle p^3$...
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Bruno J. View Post
    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 $\displaystyle p^3$...
    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)
    Last edited by Swlabr; Jul 2nd 2009 at 11:32 PM.
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    $\displaystyle G=\left \{\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z} \right \}.$
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    $\displaystyle G=\left \{\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z} \right \}.$
    That group is slightly more intuitively obvious...
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