Dihedral Group D6 is Internal Direct Product of C2 with D3

Theorem
The dihedral group $D_6$ is an internal direct product of the cyclic group $C_2$ of order $2$ and the dihedral group $D_3$:
 * $D_6 = C_2 \times D_3$

Proof
Let $G$ be defined by its group presentation:


 * $G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^{-1} }$

or:
 * $G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^5}$

Let $z$ denote $x^3$.

Then:

So $z$ commutes with $y$.

As $z$ is a power of $x$, $z$ also commutes with $x$.

Hence by definition of center:
 * $z \in \map Z G$

It follows that $\gen z$ is a normal subgroup of order $2$.

Let $N$ be the subgroup of $G$ generated by $x^2$ and $y$.

Note that:

Hence the generator of $N$ satisfies:


 * $\paren{x^2}^3 = e = y^2$

and:
 * $y x^2 y^{-1} = x^{-2}$

Let $w := x^3$.

Then $N$ is generated by $w$ and $y$ where:


 * $w^3 = 1 = y^2$

and:
 * $w y = y w^2 = y w^{-1}$

and it is seen that $N$ is isomorphic to the dihedral group $D_3$.

It is now to be shown that $G$ is an internal group direct product of $N$ and $\gen z$.

We have that $N \cap \gen z = \set e$.

From the Internal Direct Product Theorem, we need to prove only that $N$ is a normal subgroup of $G$.

We have that:
 * $\order N = 6$

where $\order N$ denotes the order of $N$.

We also have that $N$ is a subgroup of its normalizer $\map {N_G} N$.

Hence by Lagrange's Theorem:
 * $6 \divides \order {\map {N_G} N}$

where $\divides$ denotes divisibility.

Again by Lagrange's Theorem:
 * $\order {\map {N_G} N} \divides 12$

We have:

demonstrating that $x$ is conjugate to $w$.

Then:

demonstrating that $x$ is conjugate to $y$.

Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 6$.

As $6 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 12$, it follows that:
 * $\map {N_G} N = G$

and so $N$ is normal in $G$.

Thus:
 * $N$ and $\gen z$ are normal in $G$
 * $N \cap \gen z = \set e$
 * $N \gen z = G$

and it therefore follows from the Internal Direct Product Theorem that:
 * $G = C_2 \times D_3$