Is this identity for the Dihedral group correct?

Question

Let $D_n$ represents the Dihedral group with $2n$ elements, and my question(based on some physics backgrounds) is:

Does $Z_2$ a normal subgroup of $Q_8$? If it is, then is the indentity $D_2\cong Q_8/Z_2$ correct?

Here, we take $D_2=\left \{\begin{pmatrix} 1 & 0 &0 \\0 &1&0 \\0 & 0& 1\end{pmatrix}, \begin{pmatrix} 1 & 0 &0 \\0 &-1&0 \\0 & 0& -1\end{pmatrix},\begin{pmatrix} -1 & 0 &0 \\0 &1&0 \\0 & 0& -1\end{pmatrix},\begin{pmatrix}- 1 & 0 &0 \\0 &-1&0 \\0 & 0& 1\end{pmatrix} \right \}$, $Z_2=\left \{ \mathbf{I}, -\mathbf{I} \right \}$, $Q_8=\left \{ \mathbf{I}, -\mathbf{I}, e^{i\pi S_x}, e^{-i\pi S_x}, e^{i\pi S_y}, e^{-i\pi S_y}, e^{i\pi S_z}, e^{-i\pi S_z}\right \}$ with the $2\times2$ identity matrix $ \mathbf{I}$ , the three matrices $S_x=\begin{pmatrix} 0&\frac{1}{2} \\ \frac{1}{2}& 0 \end{pmatrix},S_y=\begin{pmatrix} 0&-\frac{i}{2} \\ \frac{i}{2}& 0 \end{pmatrix},S_z=\begin{pmatrix} \frac{1}{2} &0 \\ 0& - \frac{1}{2} \end{pmatrix}$ and $i=\sqrt{-1}$.

Thank you very much.

Answer 1

$D_4$ has 2 subgroups isomorphic to $Z_2$. Picture $D_4$ as the group of rotations and reflections of a square, with generator $a$ of order 4 rotating the square by 90 degrees and generator $b$ of order 2 reflecting the square over y-axis.

The subgroup generated by $b$ is not normal: observe that $aba^{-1}$ is the reflection over x-axis, which is not in the subgroup generated by $b$.

On the other hand, the other subgroup of order 2 generated by $a^2$ is in fact normal: it's a 180-degree rotation that commutes with the rest of $D_4$. And yes, $D_4/(a^2)$ is isomorphic to $D_2=Z_2\times Z_2$ because it's the only group of order 4 without elements of order 4.

Answer 2

Let $α$ denote the reflection and $β$ the rotation by $2\pi/n$. If $n=2m$, then we have a normal subgroup $\langleβ^m\rangle$ of order $2$. The other elements of order $2$ are the elements of the form $αβ^k$ for any $k$. We see that $αβ^lαβ^kαβ^l=β^{k-l}αβ^l=αβ^{2l-k}$. So in order for $\langleαβ^k\rangle$ to be normal, $2(l-k)$ must be a multiple of $n$ for each $l=0,...,n-1$. But this implies that $n=2$. In that case $D_2$ equals $C_2\times C_2$ and $D_1=C_2=D_2/C_2$ (Here $C_2$ denotes the cyclic group of order $2$).
In case $n=2m$ for $m>1$, the only normal subgroup of order $2$ is generated by $β^m$. And then $D_m\cong D_n/\langleβ^m\rangle$, since $D_n$ is the group on two generators $α,β$ under the relations $α^2=β^n=αβαβ=1$, and adding the relations $\beta^m=1$ yields $D_m$.

Answer 3

For my example, it's direct to show that $Z_2$ is a normal subgroup of $Q_8$, and $$Q_8/Z_2=\left \{ \left \{ \mathbf{I},-\mathbf{I} \right \} ,\left \{ e^{i\pi S_x},e^{-i\pi S_x} \right \},\left \{ e^{i\pi S_y},e^{-i\pi S_y} \right \},\left \{ e^{i\pi S_z},e^{-i\pi S_z} \right \}\right \}$$, where $e^{-i\pi S_\alpha } =-e^{i\pi S_\alpha }, \alpha =x,y,z $.

Furthermore, $D_2\cong Q_8/Z_2$.