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Homework Help: Abstract Algebra - Smallest subgroup of GL(n,R)

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A = [itex]\left[ \begin{array}{cccc} 1 & 1 \\ 0 & -1 \end{array} \right][/itex]

    Let B = [itex]\left[ \begin{array}{cccc} 1 & 2 \\ 0 & -1 \end{array} \right][/itex]

    Find the smallest subgroup G of GL(n,R) that contains [itex]A[/itex] and [itex]B[/itex]. Also, find the smallest subgroup H of G that contains the matrices AB and BA.

    2. Relevant equations

    GL(n,R) is the group of all n x n invertible matrices


    [itex]f(a*b) = f(a) * f(b)[/itex] for some operation [itex]*[/itex]

    3. The attempt at a solution

    I need a hint to start (don't solve the whole thing, please just help me start it).

    So far, I can see that [itex]A=[/itex][itex]A^{-1}[/itex] and [itex]B=[/itex][itex]B^{-1}[/itex]

    Also, for the second part [itex]AB\neq BA[/itex] since [itex]AB=\left[ \begin{array}{cccc} 1 & 1 \\ 0 & 1 \end{array} \right][/itex] and [itex]BA=\left[ \begin{array}{cccc} 1 & -1 \\ 0 & 1 \end{array} \right][/itex], also, [itex]\left(AB\right)^{-1}=\left[ \begin{array}{cccc} 1 & -1 \\ 0 & 1 \end{array} \right]=BA[/itex], and, [itex]\left(BA\right)^{-1}=\left[ \begin{array}{cccc} 1 & 1 \\ 0 & 1 \end{array} \right]=AB[/itex]

    To show that something is a subgroup, I need to show that it is closed under the operation (matrix multiplication in the second part), inverses, contains the identity element, and has the homomorphism property

    Can I define an operation that just gives the inverse of the matrix? (meaning I think the set {I, A, B} under that operation would be the smallest subgroup), and a similar group can be taken for the second part, just having it contain {I, AB, BA}.

    Or, would the following work: The set {I, A, B, AB, BA} under matrix multiplication?
    with this:
    For any two elements, say [itex]X[/itex] and [itex]Y[/itex] of the set, [itex]X*Y[/itex] is in the set as well (given what I've shown above).

    [itex]A*(B*BA)=(A*B)*BA[/itex], and this seems to work for all such combinations, so associativity holds.

    Inverses hold given what I've shown above, and the identity is there by definition.
    Last edited: Feb 14, 2012
  2. jcsd
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