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)
\begin{equation} <-1.i,j,k| (-1)^2=1, i^2=j^2=k^2=ijk=-1> \end{equation}

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

(2)
\begin{equation} <-1.i,j,k| (-1)^2=1, i^2=j^2=k^2=ijk=-1> \end{equation}

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)
\begin{equation} ij=k \end{equation}
(9)
\begin{equation} jk=i \end{equation}
(11)
\begin{equation} ki=j \end{equation}
(13)
\begin{equation} ik=-j \end{equation}
(15)
\begin{equation} kj=-i \end{equation}
(17)
\begin{equation} ji=-k \end{equation}

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

Inverses:

(22)
\begin{equation} j(-j)=1 \end{equation}
(24)
\begin{equation} i(-i)=1 \end{equation}
(26)
\begin{equation} k(-k)=1 \end{equation}
(28)
\begin{equation} (-1)(-1)=1 \end{equation}

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

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