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Math Help - Abstract Algebra

  1. #1
    anythink25
    Guest

    Abstract Algebra

    The quaternion group Q may be defined by generators and relations as follows:
    Q= <x,y | x^4 =1,y^2=x^2,yxy^-1= x^-1>
    for another description let i= squareroot (-1) and take subgroup of 2x2 matrix generated by x= martix going across- 0,-1,1,0 and y = i,0,o,-i

    A- explain why every element of Q may be written uniquily in the standard form x^sy^t with s=0,1,2,3 and t= 0,1 hence |Q|= 8

    b- verify that the order of y is 4 and xyx^-1 = y^-1
    c-complete the following multiplication table representing each result in standard form
    x^s (x^s) y
    x^s
    (x^s)y

    d- find all elements of order 2 in Q
    e- find the center of Q
    please explain as much as you can thanks
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  2. #2
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    (A)
    x has order 4, so 1,x,x^2,x^3\;(=x^{-1}) are distinct elements; they are also distinct from y (otherwise Q would be cyclic). So we have five distinct elements. And since y^2=x^2, it follows that Q has at most eight elements: 1,x,x^2,x^3,y,xy,x^2y,x^3y. All you need to do is to show that the last three of these elements are distinct from the first five and from each other. For example, for xy:
    • If xy=1, then y=x^{-1}=x^3 (contradiction). Hence xy\ne1.
    • If xy=x, then y=1 (contradiction). Hence xy\ne x.
    • If xy=x^2, then y=x (contradiction). Hence xy\ne x^2.
    • If xy=x^3, then y=x^2 (contradiction). Hence xy\ne x^3.
    • If xy=y, then x=1 (contradiction). Hence xy\ne y.

    Show similarly that x^2y,x^3y\not\in\{1,x,x^2,x^3,y\}, then show that xy\ne x^2y,x^2y\ne x^3y,x^3y\ne xy.

    (B)
    We have y\ne1,\ y^2=x^2\ne1,\ y^3=(y^2)y=x^2y\ne1 and y^4=(y^2)^2=(x^2)^2=x^4=1.

    xyx^{-1}=xy(yxy^{-1}) … and the rest is straightforward.

    (C)
    For any k\in\mathbb{Z}, x^k=x^{k'} where k'\in\{0,1,2,3\} and k\equiv k'\pmod{4}.

    (D)
    x^2 is the only element of order 2. This is just a matter of checking one by one.

    (E)
    We have yx=yxy^{-1}y=x^{-1}y\ne xy; hence x and y do not belong to the centre, and neither do their inverses x^3 and x^2y. Also (xy)y=xy^2=x^3 but yxy=yx(y^{-1}y)y=(yxy^{-1})y^2=x^{-1}x^2=x\ne (xy)y. Hence, neither xy nor (xy)^{-1}=x^3y belongs to the centre. That leaves x^2. Simple checking should verify that x^2 does commute with every other element in Q. Hence \mathrm{Z}(Q)=\{1,x^2\}.
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