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Thread: Abstract Algebra

  1. #1

    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

    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
    Dec 2007
    x has order 4, so $\displaystyle 1,x,x^2,x^3\;(=x^{-1})$ are distinct elements; they are also distinct from y (otherwise Q would be cyclic). So we have five distinct elements. And since $\displaystyle y^2=x^2$, it follows that Q has at most eight elements: $\displaystyle 1,x,x^2,x^3,y,xy,x^2y,x^3y$. 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 $\displaystyle xy=1$, then $\displaystyle y=x^{-1}=x^3$ (contradiction). Hence $\displaystyle xy\ne1$.
    • If $\displaystyle xy=x$, then $\displaystyle y=1$ (contradiction). Hence $\displaystyle xy\ne x$.
    • If $\displaystyle xy=x^2$, then $\displaystyle y=x$ (contradiction). Hence $\displaystyle xy\ne x^2$.
    • If $\displaystyle xy=x^3$, then $\displaystyle y=x^2$ (contradiction). Hence $\displaystyle xy\ne x^3$.
    • If $\displaystyle xy=y$, then $\displaystyle x=1$ (contradiction). Hence $\displaystyle xy\ne y$.

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

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

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

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

    $\displaystyle x^2$ is the only element of order 2. This is just a matter of checking one by one.

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