# Math Help - Abstract Algebra

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
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

2. (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 $\mathrm{Z}(Q)=\{1,x^2\}$.