Group Homomorphism/Examples/Mapping from D3 to Parity Group

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Examples of Group Homomorphisms

Let $D_3$ denote the symmetry group of the equilateral triangle:

\(\ds e\) \(:\) \(\ds \paren A \paren B \paren C\) Identity mapping
\(\ds p\) \(:\) \(\ds \paren {ABC}\) Rotation of $120 \degrees$ anticlockwise about center
\(\ds q\) \(:\) \(\ds \paren {ACB}\) Rotation of $120 \degrees$ clockwise about center
\(\ds r\) \(:\) \(\ds \paren {BC}\) Reflection in line $r$
\(\ds s\) \(:\) \(\ds \paren {AC}\) Reflection in line $s$
\(\ds t\) \(:\) \(\ds \paren {AB}\) Reflection in line $t$


SymmetryGroupEqTriangle.png


Let $G$ denote the parity group, defined as:

$\struct {\set {1, -1}, \times}$

where $\times$ denotes conventional multiplication.


Let $\theta: D_3 \to G$ be the mapping defined as:

$\forall x \in D_3: \map \theta x = \begin{cases} 1 & : \text{$x$ is a rotation} \\ -1 & : \text{$x$ is a reflection} \end{cases}$


Then $\theta$ is a (group) homomorphism, where:

\(\ds \map \ker \theta\) \(=\) \(\ds \set {e, p, q}\)
\(\ds \Img \theta\) \(=\) \(\ds G\)


Proof

Let $x, y \in D_3$.

Let $P$ denote the subset of $D_3$ consisting of the rotations:

$P := \set {e, p, q}$

Let $R$ denote the subset of $D_3$ consisting of the reflections:

$R := \set {r, s, t}$

It is noted that $e$ is classified as a rotation of zero angle.


The Cayley table for $D_3$ is:

$\begin{array}{c|ccc|ccc} \circ & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ \hline r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$


Direct reference to this Cayley table gives us:

\(\ds \forall x, y \in P: \, \) \(\ds x \circ y\) \(\in\) \(\ds P\)
\(\ds \forall x, y \in R: \, \) \(\ds x \circ y\) \(\in\) \(\ds P\)
\(\ds \forall x \in P, y \in R: \, \) \(\ds x \circ y\) \(\in\) \(\ds R\)
\(\ds \forall x \in R, y \in P: \, \) \(\ds x \circ y\) \(\in\) \(\ds R\)


Thus we have:

\(\ds \forall x, y \in P: \, \) \(\ds \map \phi x \times \map \phi y\) \(=\) \(\ds 1 \times 1\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\)


\(\ds \forall x, y \in R: \, \) \(\ds \map \phi x \times \map \phi y\) \(=\) \(\ds \paren {-1} \times \paren {-1}\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\)


\(\ds \forall x \in P, y \in R: \, \) \(\ds \map \phi x \times \map \phi y\) \(=\) \(\ds 1 \times \paren {-1}\)
\(\ds \) \(=\) \(\ds -1\)
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\)


\(\ds \forall x \in R, y \in P: \, \) \(\ds \map \phi x \times \map \phi y\) \(=\) \(\ds \paren {-1} \times 1\)
\(\ds \) \(=\) \(\ds -1\)
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\)


By definition, $\map \ker \phi$ is the set of all elements of $D_3$ which map to $1$.

Hence by definition:

$\map \ker \phi = \set {e, p, q}$


Also by definition, $\Img \phi$ is the set of all elements of $G$ which are mappped to by $\phi$.

Hence by definition:

$\Img \phi = \set {1, -1}$


Hence the result.

$\blacksquare$


Sources

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