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Math Help - matrix groups

  1. #1
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    matrix groups

    A= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)
    , B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)
    \\
    Show that \langle A, B \rangle is subgroup of GL_2(\mathbb{C}). And Show that \langle A, B \rangle generated by A and B, and order of \langle A, B \rangle is 8 ?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: matrix groups

    Quote Originally Posted by burak100 View Post
    A= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)
    , B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)
    \\
    Show that \langle A, B \rangle is subgroup of GL_2(\mathbb{C}). And Show that \langle A, B \rangle generated by A and B, and order of \langle A, B \rangle is 8 ?
    Note that A,B satisfy the relations A^4=B^4=1 and AB=A^{-2}BA and so you can easily verify that the map f:Q_8\to\text{GL}_2(\mathbb{C}) (where Q_8 is the Quaternion group) defined by i\mapsto A and j\mapsto B is a homomorphism, and since it maps generators to generators we see that it surjects onto \langle A,B\rangle, but clearly  \text{im }f\leqslant AB and so \text{im }f=\langle A,B\rangle. Now, since \langle A,B\rangle can be covered by Q_8 we know that \left|\langle A,B\rangle\right|\leqslant 8 but since \left|\langle A,B\rangle\right|\mid |Q_8|=8 you must have that \left|\langle A,B\rangle\right\=1,2,4,8. But, since (as can easily be checked) I,-I,A,-A,B are 5 distinct elements of \langle A,B\rangle we may conclude that \left|\langle A,B\rangle\right|=8 and moreover that f is an isomorphism (since a finite surjection between two sets of equal cardinality must be a bijection).
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  3. #3
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    Re: matrix groups

    this is for order of <A, B> right?
    How can I show that <A, B> is generated by A and B
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: matrix groups

    Well, I don't know what notation you are using but the conventional notation is \langle S\rangle is the subgroup generated by the set S. So, by DEFINITION A,B generate \langle A,B\rangle.
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    Re: matrix groups

    Actually , in the question doesn't define what is <A, B> but , I think that means; check that the group generated by all possible products of A and B is a subgroup of order 8????

    because if we think that <A, B> is just an inner product then <A, B>= \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix}\right) and it is just an element of GL_2(C) right? not a subgroup....?
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  6. #6
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    Re: matrix groups

    Perhaps try to just generate the subgroup with A and not B in order to arrive at the contradiction that you need both in order to generate the subgroup? Then try just B and not A?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Re: matrix groups

    Quote Originally Posted by burak100 View Post
    Actually , in the question doesn't define what is <A, B> but , I think that means; check that the group generated by all possible products of A and B is a subgroup of order 8????

    because if we think that <A, B> is just an inner product then <A, B>= \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix}\right) and it is just an element of GL_2(C) right? not a subgroup....?
    This doesn't make sense. It's conceivable that they have defined the group \langle S\rangle to be the set of all words (I assume you know what that means) in S but this is equivalent to generated.
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  8. #8
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    Re: matrix groups

    I, A, B, AB, BA, AAB, AAA, BAA these are the possibilities with using the fact A^4=I and B^2=I and AB=-BA
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  9. #9
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    Re: matrix groups

    i am confused when reading this thread. for example i get that B^2 = I, so that B is of order 2. furthermore i agree with the OP, and disagree with drexel28 that: AB = -BA, rather than AB = A^-2BA.

    first, let's agree on a standard way to write the possible products:

    since AB = -BA, whenever we get a B before an A, let's replace the BA with -AB. this means we can always put the A's first.

    also, note that A^2 = -I, so A^3 = -A. so let's agree to write all instances of -A as A^3, and when the powers of A get greater than 4,

    we can reduce them, since A^4 = I.

    now, we know I, A, A^2 and A^3 are all different, so that's 4 elements. since B is of order 2, the only possible products left are:

    B, AB, A^2B and A^3B. this tells us we have 8 elements in the set of all possible products of A's and B's.

    to show this is a subgroup of GL2(C), we need only show closure, as this is a finite set.

    but it should be clear that we can write [(A^j1)(B^k1)][(A^j2)(B^k2)], j1,j2 = 0,1,2 or 3 and k1,k2 = 0 or 1 (here i am using the convention that A^0 = B^0 = I).

    as one of our 8 products above by pulling the A^j2 term in front of the B^k1 term, using BA = A^3B (repeatedly, if necessary).

    for example: (A^3B)(A^2B) = (A^3)(BA)(AB) = (A^3)(A^3B)(AB) = (A^6B)AB = (A^2B)AB

    = A^2(BA)B = A^2(A^3B)B = A^5B^2 = A.

    alternatively, we can note that the rules A^4 = B^2 = I, BA = A^3B = A^-1B are the same generating relations as the dihedral group of order 8,

    with <A> being the rotation group, and B a generating reflection.
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