Question

Prove that the dihedral group ${\mathit{D}}_{\mathbf{6}}$ is isomorphic to ${\mathit{S}}_{\mathbf{3}}\mathbf{\times }{\mathrm{\mathbb{Z}}}_{\mathbf{2}}$.

Short answer

It is proved that, $D_6$ is isomorphic to $S_3\times {\mathrm{\mathbb{Z}}}_2$.

Step-by-step solution

Step 1:To show f  is homomorphism

Define a mapping $f:D_6\to S_3\times {\mathrm{\mathbb{Z}}}_2$by $\left(r^i{d}^j\right)\mapsto \left({\left(123\right)}^i{\left(12\right)}^j,\left[i\right]\right)$.

Now, we prove f is a homomorphism as follows:

$\begin{array}{rcl}f\left(\left(r^i{d}^j\right)\left(r^m{d}^n\right)\right)& =& f\left(r^{i+{\left(-1\right)}^{J}m}{d}^{j+n}\right)\\ & =& \left({\left(123\right)}^{i+{\left(-1\right)}^{J}m}{\left(12\right)}^{j+n},\left[i+m\right]\right)\\ & =& \left({\left(123\right)}^i{\left(12\right)}^j,\left[i\right]\right)\left({\left(123\right)}^m{\left(12\right)}^n,\left[m\right]\right)\\ & =& f\left(r^i{d}^j\right)f\left(r^m{d}^n\right)\end{array}$

$\therefore f\left(\left(r^i{d}^j\right)\left(r^m{d}^n\right)\right)=f\left(r^i{d}^j\right)f\left(r^m{d}^n\right)$

Hence, f is a homomorphism.

To prove D6  is isomorphic to S3×ℤ2

If $f\left(r^i{d}^j\right)=\left(1,\left[0\right]\right)$, then ${\left(123\right)}^i{\left(12\right)}^j=1$and $\left[i\right]=0$.

Simplify ${\left(123\right)}^i{\left(12\right)}^j=1$ as:

${\left(123\right)}^i{\left(12\right)}^j=1$

$\Rightarrow 2|j$and $3|i$

And$\left[i\right]=0\Rightarrow 2|i$

Therefore, $6|i$.

Thus, $r^i{d}^j=e$.

Therefore, f is injective.

Since $\left|D_6\right|=12$and $\left|S_3\times {\mathrm{\mathbb{Z}}}_2\right|=1$, we can say that f is an isomorphism.

Hence, $D_6$ is isomorphic to $S_3\times {\mathrm{\mathbb{Z}}}_2$.