Theorem: Dihedral Group D6 is Internal Direct Product of C2 with D3

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:
{{begin-eqn}}
{{eqn | l = y z y^{-1}
      | r = y x^3 y^{-1}
      | c = 
}}
{{eqn | r = \paren {y x y^{-1} }^3
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {x^5}^3
      | c = {{Defof|Group Presentation}}
}}
{{eqn | r = x^{15}
      | c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = x^{12} x^3
      | c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = e \cdot x^3
      | c = {{Defof|Group Presentation}}
}}
{{eqn | r = x^3
      | c = {{Group-axiom|2}}
}}
{{eqn | r = z
      | c = Definition of $z$
}}
{{eqn | ll= \leadsto
      | l = y z
      | r = z y
      | c = Product of both sides with $y$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = y x^2 y^{-1}
      | r = \paren {y x y^{-1} }^2
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {x^5}^2
      | c = {{Defof|Group Presentation}}
}}
{{eqn | r = x^{10}
      | c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = x^{-2}
      | c = as $x^{10} = e$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = x w x^{-1}
      | r = x x^2 x^{-1}
      | c = Definition of $w$
}}
{{eqn | r = x^2
      | c = 
}}
{{eqn | r = w
      | c = 
}}
{{end-eqn}}
demonstrating that $x$ is conjugate to $w$.
Then:
{{begin-eqn}}
{{eqn | l = x y x^{-1}
      | r = x y^{-1} x^{-1}
      | c = as $y^2 = e$
}}
{{eqn | r = y x^5 x^{-1}
      | c = {{Defof|Group Presentation}}: $y x y^{-1} = x^5$
}}
{{eqn | r = y x^2
      | c = 
}}
{{eqn | r = y w
      | c = 
}}
{{eqn | o = \in
      | r = N
      | c = 
}}
{{end-eqn}}
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:
:$G = C_2 \times D_3$
{{qed}}