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Math Help - Algebra, Problems For Fun (30)

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    Algebra, Problems For Fun (30)

    True of false: For every integer n \geq 2, there exists a non-abelian group of order n^3.
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    MHF Contributor Bruno J.'s Avatar
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    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...
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    MHF Contributor Swlabr's Avatar
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    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)
    Last edited by Swlabr; July 2nd 2009 at 11:32 PM.
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    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
    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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