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Problems regarding group presentations and submodules

  1. May 25, 2010 #1
    Hi! I'm studying for an exam in group- and ring theory, and I have some questions about two problems that I have not managed to solve. I would greatly appreciate help.

    Problem 1. Determine the order of the group G with the presentation [tex](a,b \big\vert\: a^{6} = 1, b^{2} = a^{3}, ba = a^{-1}b)[/tex].

    For the first problem, the relation [tex]ba = a^{-1}b[/tex] enables us to conclude that the order of [tex]G[/tex] is less than- or equal to 24 (we can collect all the a's to the left, and the largest possible orders for a and b are 6 and 4, respectively. Moreover, using the relation [tex]b^{2} = a^{3}[/tex] we can write all elements as [tex]a^{k}b^{j}[/tex] as [tex]k[/tex] lies in the range 0 to 5, and j is either 0 or 1. This shows that the group has order less than or equal to 12. Also, the group has order [tex]\geq 6[/tex], simply since we can construct a group of that order satisfying the relations (assume a is of order 3). My problem is that I cannot find clear arguments for why the group should have order 12, which I believe. Could I in some nice way just prove that for the collection of symbols [tex]a^{k}b^{j}[/tex] above, the group axioms hold? Does there perhaps exist a theorem concerning presentations of this type?

    Problem 2. Let M be the [tex]\mathbb{C}[x][/tex]-module [tex]\mathbb{C}^{3}[/tex] where [tex]\mathbb{C}[/tex] acts naturally and x acts via [tex]x \cdot a = T \cdot a[/tex], for elements [tex]a \in \mathbb{C}[/tex] and a given linear transformation T. How can one determine all submodules of M?

    For this problem less of an idea. I understand that the actions of [tex]\mathbb{C}[/tex] and x induces an action of any complec polynomial, and the submodules are clearily the subspaces of [tex]\mathbb{C}^{3}[/tex] stable under the linear map T. I guess some stable subspaces would be those spanned by eigenvectors for T. Are these all? I haven't thought alot about this, but I'm running out of time, and would thus be very thankful for some help.

    Thanks in advance!
  2. jcsd
  3. May 26, 2010 #2
    Because the presentation is pretty nice, it's not too hard to show that you get a group by having elements of the form a^i b^j, i=0,...5, j=0,1 and specifying the multiplication rule. Alternatively, a semi-direct product of Z_3 and Z_4 has that presentation: http://shell.cas.usf.edu/~wclark/algctlg/small_groups.html
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