Daniil Perebeynos

I am working with the Quaternion group, it is non-abelian group of the order 8.
Elements of the Quaternion group are

(1)
$$<-1.i,j,k| (-1)^2=1, i^2=j^2=k^2=ijk=-1>$$

Quaternial group can be represented as a subset of general linear group GL2 (C).
Then

(2)
$$<-1.i,j,k| (-1)^2=1, i^2=j^2=k^2=ijk=-1>$$

Can be represented in matrix form as:

(3)
\begin{align} $1= \left( {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right)$ $-1= \left( {\begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} } \right)$ \end{align}
(4)
\begin{align} $i= \left( {\begin{array}{cc} \sqrt{-1} & 0 \\ 0 & - \sqrt{-1} \\ \end{array} } \right)$ $-i= \left( {\begin{array}{cc} - \sqrt{-1} & 0 \\ 0 & \sqrt{-1} \\ \end{array} } \right)$ \end{align}
(5)
\begin{align} $j= \left( {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right)$ $-j= \left( {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right)$ \end{align}
(6)
\begin{align} $k= \left( {\begin{array}{cc} 0 & \sqrt{-1} \\ \sqrt{-1} & 0 \\ \end{array} } \right)$ $-k= \left( {\begin{array}{cc} 0 & - \sqrt{-1} \\ - \sqrt{-1} & 0 \\ \end{array} } \right)$ \end{align}

To check if subset would give us subgroup, we need to check for clousure, identity element and inverses. As we know assosiativity comes from GL2(C)

Clousure:

(7)
$$ij=k$$
(9)
$$jk=i$$
(11)
$$ki=j$$
(13)
$$ik=-j$$
(15)
$$kj=-i$$
(17)
$$ji=-k$$

Identity element:(19)
\begin{align} $1= \left( {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right)$ \end{align}

Inverses:

(22)
$$j(-j)=1$$
(24)
$$i(-i)=1$$
(26)
$$k(-k)=1$$
(28)
$$(-1)(-1)=1$$

Subgroups, Left and Right cosets of Quaternial group

Let H be a subgroup of my group:
H={1,-1,i.-i}

Then left cosets will be:
kH={k,-k,-j,j}
Then right cosets will be
Hk={k,-k,-j,j}

Since kH=Hk and kM=Mk, iM=Mi, jM=Mj then by definition we say H and M are normal subgroups of our group Q8

page revision: 88, last edited: 09 Dec 2009 18:43