Dihedral Group D6 is Internal Direct Product of C2 with D3

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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:

\(\ds y z y^{-1}\) \(=\) \(\ds y x^3 y^{-1}\)
\(\ds \) \(=\) \(\ds \paren {y x y^{-1} }^3\) Power of Conjugate equals Conjugate of Power
\(\ds \) \(=\) \(\ds \paren {x^5}^3\) from Group Presentation
\(\ds \) \(=\) \(\ds x^{15}\) Powers of Group Elements: Product of Indices
\(\ds \) \(=\) \(\ds x^{12} x^3\) Powers of Group Elements: Sum of Indices
\(\ds \) \(=\) \(\ds e \cdot x^3\) from Group Presentation
\(\ds \) \(=\) \(\ds x^3\) Group Axiom $\text G 2$: Existence of Identity Element
\(\ds \) \(=\) \(\ds z\) Definition of $z$
\(\ds \leadsto \ \ \) \(\ds y z\) \(=\) \(\ds z y\) Product of both sides with $y$

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:

\(\ds y x^2 y^{-1}\) \(=\) \(\ds \paren {y x y^{-1} }^2\) Power of Conjugate equals Conjugate of Power
\(\ds \) \(=\) \(\ds \paren {x^5}^2\) from Group Presentation
\(\ds \) \(=\) \(\ds x^{10}\) Powers of Group Elements: Product of Indices
\(\ds \) \(=\) \(\ds x^{-2}\) as $x^{10} = e$

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:

\(\ds x w x^{-1}\) \(=\) \(\ds x x^2 x^{-1}\) Definition of $w$
\(\ds \) \(=\) \(\ds x^2\)
\(\ds \) \(=\) \(\ds w\)

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


Then:

\(\ds x y x^{-1}\) \(=\) \(\ds x y^{-1} x^{-1}\) as $y^2 = e$
\(\ds \) \(=\) \(\ds y x^5 x^{-1}\) from Group Presentation: $y x y^{-1} = x^5$
\(\ds \) \(=\) \(\ds y x^2\)
\(\ds \) \(=\) \(\ds y w\)
\(\ds \) \(\in\) \(\ds N\)

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:

$G = C_2 \times D_3$

$\blacksquare$


Sources

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