# Math 410 Final Project: Representing Dihedral Groups as a Group of Permutations, by Natalie Abreu

The goal of this article is to examine how we can represent dihedral groups with groups of permutations. In particular, we will look at mapping $D_{2n}$ to a subgroup of $S_{n}$, and generalize this to mapping $D_{2n}$ to a subgroup of $S_{2n}$. We will start by considering how to represent the symmetries of an equilateral triangle ($D_{6}$) to permutations of three characters and six characters (subgroups of $S_3$ and $S_6$, respectively), and use this to generalize to regular n-gons. This example serves as a nice illustration of Cayley's Theorem, which states that every group is isomorphic to a subgroup of the symmetric group $S_n$ for some n. Additionally, I have created animations to illustrate some of the concepts.

## Representing $D_{6}$ with $S_{3}$

In this section we will examine how we can represent the symmetries of an equilateral triangle with permutations of three characters. A natural place to start might be to associate each vertex of the triangle with a character.

> **Proposition A.1**
> Let $D_{6}$ = $<r,s: r^3 = s^2 = 1, sr= rs^2>$. Then $\phi: D_{6} \rightarrow S_3$ where $\phi: r \mapsto (1 2 3), s \mapsto (2 3)$ induces an isomorphism from $D_6$ to $S_3$.

The following illustrates this mapping:

| <b>Visual representation of $\phi$ </b>|

*Proof.* To prove that $\phi$ is an isomorphism, we must show the following
$\bullet$ $\phi$ is an group homomorphism
$\bullet$ $\phi$ is injective and surjective

To show $\phi$ is a group homomorphism, we want to show that it respects both group structures. Formally, $\phi(g1*g2) = \phi(g1) * \phi(g2)$ for all $g1$, $g2$ in $D_6$. We can write the multiplication table to verify this: (Note that rows are permuted in order to have the identity element along the diagonal, and that $e$ is used to refer to the identity of $D_6$ and $Id$ is used to refer to the identity of $S_3$ to avoid confusion).

$$ \begin{array}{c|cccccc}
& e & r & r^2 & s & sr & sr^2 \\ \hline
~e~ & e & r & r^2 & s & sr & sr^2\\
~r^2~ & r^2 & e & r & sr^2 & s & sr \\
~r~ & r & r^2 & e & sr & sr^2 & s \\
~s~ & s & sr^2 & sr & e & r^2 & r \\
~sr~ & sr & s & sr^2 & r & e & r^2 \\
~sr^2~ & sr^2 & sr & s & r^2 & r & e \\
\end {array} $$

This aligns with multiplication of elements in $S_3$:
$$ \begin{array}{c|cccccc}
& Id & (1 2 3) & (1 3 2) & (2 3) & (1 3) & (1 2) \\ \hline
~Id~ & Id & (1 2 3) & (1 3 2) & (2 3) & (1 3) & (1 2)\\
~(1 3 2)~ & (1 3 2) & Id & (1 2 3) & (1 2) & (2 3) & (1 3) \\
~(1 2 3)~ & (1 2 3) & (1 3 2) & Id & (1 3) & (1 2) & (2 3) \\
~(2 3)~ & (2 3) & (1 2) & (1 3) & Id & (1 3 2) & (1 2 3) \\
~(1 3)~ & (1 3) & (2 3) & (1 2) & (1 2 3) & Id & (1 3 2) \\
~(1 2)~ & (1 2) & (1 3) & (2 3) & (1 3 2) & (1 2 3) & Id \\
\end {array} $$

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| <b>Example of the group homomorphism property of $\phi$</b> |

Since $|D_{6}| = |S_3| = 6$, it suffices to show injectivity or surjectivity to show $\phi$ is an isomorphism. This mapping is clearly injective, as each element in $D_6$ corresponds to a different element in $S_3$.

## Representing $D_6$ with Subgroups of $S_6$

We can also consider how we might extend this representation to permutations of six characters ($S_6$). Since $|S_6|$ > $|D_6|$, we must select a subset of elements of $S_6$ to use for our mapping, as well as make sure that this subset is a subgroup of $S_6$. In this section, we will examine how we might do this in two essentially different ways.

### Method 1

A natural method of doing so would simply be to preserve our prior map and fix the remaining characters.

> **Proposition B.1**
> Let $D_{6}$ = $<r,s: r^3 = s^2 = 1, sr= rs^2>$. Then $\phi: D_{6} \rightarrow S_6$ where $\phi: r \mapsto (1 2 3), s \mapsto (2 3)$ induces an homomorphism from $D_6$ to $S_6$.

The same multiplication tables from above verify that this is a group homomorphism.
More specifically though, we want to claim that the elements $\{(Id), (1 2 3), (1 3 2), (2 3), (1 3), (1 2)\} \subset S_6$ form a subgroup $H$, and the mapping $\phi: D_{6} \rightarrow H$ where $\phi: r \mapsto (1 2 3), s \mapsto (2 3)$ induces an *isomorphism* from $D_6$ to $H$.

That these elements form a subgroup of $S_6$ isn't too difficult to show:
The identity of $H$ is the identity of $S_6$, and we retain associativity from $S_6$. The multiplication table additionally tells us that the subgroup is closed (letters that we wanted to fix remain fixed in any multiplication of elements), and that each element has an inverse in $H$.

The mapping is clearly still injective, and as $|D_{6}| = |H| = 6$, we can conclude:

> **Proposition B.2**
> Let $D_{6}$ = $<r,s: r^3 = s^2 = 1, sr= rs^2>$ and $H = \{Id, (1 2 3), (1 3 2), (2 3), (1 3), (1 2)\} \subset S_6$. Then $\phi: D_{6} \rightarrow H$ where $\phi: r \mapsto (1 2 3), s \mapsto (2 3)$ induces an isomorphism from $D_6$ to $H$.

### Method 2

Another natural but slightly more involved method of representing $D_6$ with a group of permutations of six letters might be to consider how we might associate the edges with characters as well as the vertices. We can then extend our prior mapping as follows:

> **Proposition B.3**
> Let $D_{6}$ = $<r,s: r^3 = s^2 = 1, sr= rs^2>$. Then $\phi: D_{6} \rightarrow S_6$ where $\phi: r \mapsto (1 2 3)(4 5 6), s \mapsto (2 3)(4 6)$ induces an homomorphism from $D_6$ to $S_6$.

| <b>Visual representation of $\phi$ </b>|

Notice that if we preserve how $\phi$ affects the characters that represent vertices, then we have already induced how $\phi$ must map the characters that represent the edges.

We again can modify this to say that these elements of $S_6$ form a subgroup H, and this mapping from $D_6$ onto H is an isomorphism:
> **Proposition B.4**
> Let $D_{6}$ = $<r,s: r^3 = s^2 = 1, sr= rs^2>$ and $H = \{Id, (1 2 3)(4 5 6), (1 3 2)(4 6 5), (2 3)(4 6), (1 3)(4 5), (1 2)(5 6)\} \subset S_6$. Then $\phi: D_{6} \rightarrow H$ where $\phi: r \mapsto (1 2 3)(4 5 6), s \mapsto (2 3)(4 6)$ induces an isomorphism from $D_6$ to $H$.

*Proof.* We want to show that $H$ is in fact a subgroup, that $\phi$ is a group homomorphism, and that $\phi$ is injective. For the first two criteria, we can again use multiplication tables.

$$ \begin{array}{c|cccccc}
& Id & (1 2 3)(4 5 6) & (1 3 2)(4 6 5) & (2 3)(4 6) & (1 3)(4 5) & (1 2)(5 6) \\ \hline
~Id~ & Id & (1 2 3)(4 6 5) & (1 3 2)(4 6 5) & (2 3)(4 6) & (1 3)(4 5) & (1 2)(5 6)\\
~(1 3 2)(4 6 5)~ & (1 3 2)(4 6 5) & Id & (1 2 3)(4 5 6) & (1 2)(5 6) & (2 3)(4 6) & (1 3)(4 5) \\
~(1 2 3)(4 5 6)~ & (1 2 3)(4 5 6) & (1 3 2)(4 6 5) & Id & (1 3)(4 5) & (1 2)(5 6) & (2 3)(4 6) \\
~(2 3)(4 6)~ & (2 3)(4 6) & (1 2)(5 6) & (1 3)(4 5) & Id & (1 3 2)(4 6 5) & (1 2 3)(4 5 6) \\
~(1 3)(4 5)~ & (1 3)(4 5) & (2 3)(4 6) & (1 2)(5 6) & (1 2 3)(4 5 6) & Id & (1 3 2)(4 6 5) \\
~(1 2)(5 6)~ & (1 2)(5 6) & (1 3)(4 5) & (2 3)(4 6) & (1 3 2)(4 6 5) & (1 2 3)(4 5 6) & Id \\
\end {array} $$

This shows that $H$ is closed under multiplication, and that each element has an inverse in $H$. We additionally have that $1_H = 1_{S6}$ and we have associativity given that $S_6$ is associative, so we can conclude that $H$ is a subgroup.
The group homomorphism property $\phi(g1*g2) = \phi(g1) * \phi(g2)$ for all $g1$, $g2$ in $D_6$ holds as well, as we can check that this multiplication table aligns with the multiplication table of $D_6$.

| <b>Example of the group homomorphism property of $\phi$ </b>|

## Generalizing to $D_{2n}$

We have now examined how to represent the symmetries of an equilateral triangle using groups of permutations of three and six characters. How can we generalize this to symmetries of a regular n-gon?

First we can consider how to generalize the three character case ($D_6 \rightarrow S_3$). Let $D_{2n} = <r,s : r^n = s^2 = 1, sr=r^{n-1}s>$. We can construct a similar homomorphism $D_{2n} \rightarrow S_n$ by "assigning" each vertex a character, such that r maps to an n-cycle of those characters corresponding to rotating the vertices, and s maps to a sequence of transpositions. We can choose s to be any sequence of transpositions that corresponds to flipping the n-gon over an axis through a vertex. (If n is even, we could also choose a reflection over an axis through a pair of parallel edges). Note that while the mapping we constructed from $D_6$ to $S_3$ was an isomorphism, this does not hold for general $D_{2n} \rightarrow S_n$, as $S_n$ has order $n!$ which is greater than $2n$ for $n > 3$. So when we generalize to regular n-gons, we can instead construct an isomorphism from $D_{2n}$ to a *subgroup* of $S_n$.

| <b>Generaling to regular n-gons$ </b>|

Extending our mapping to $S_{2n}$ also follows the same idea as in our example with $D_6$. We can extend our mapping using Method 1 by retaining our mapping $D_{2n} \rightarrow S_{n}$ and simply fixing the remaining $n$ characters. In fact, this method works for mapping $D_{2n}$ to a subgroup of $S_{n'}$ for any $n' > n$.
We can extend using Method 2 by similarly mapping the remaining n characters to the edges of the n-gon. If we have characters $1, 2, ..., n$ corresponding to vertices, and characters $n+1, n+2, ..., 2n$ corresponding to edges, we get the resulting mapping where $r \mapsto (1, 2, ..., n)(n+1, n+2, ..., 2n)$, and $s \mapsto (2 \: n)(3 \: n-1)...(n+1 \: 2n)(n+1 \: 2n-1)...$. Just as with $D_6$, this gives us two methods of mapping the symmetries of a regular n-gon to a subgroup of permutations of $2n$ characters.

## References

Birkhoff, G., & Mac Lane, S. (1997). A Survey of Modern Algebra (1st ed.). A K Peters/CRC Press. https://doi-org.libproxy1.usc.edu/10.1201/9781315275499