# Constructing dihedral group:D8 from its presentation

The goal of this article is to use the presentation of dihedral group:D8 as an example illustrating how to deal with presentations, convert them into multiplication tables, and deduce basic facts about the group.

In this article, we denote by $e$ the identity element of the group.

We consider the presentation:

$\langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$

Note that due to the relation $x^2 = e$, we get $x = x^{-1}$, and an equivalent presentation is:

$\langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$

And yet another is:

$\langle a,x \mid a^4 = x^2 = xaxa = e \rangle$

## What do we care about when dealing with presentations

### Bounding from above

The group is generated by elements $a$ and $x$ and thus every element can be written as a word involving $a,x$ (possibly with negative exponents) such as $xaaaxxxa^{-1}xaxax^{-1}$. Using the shorthand for words, we can rewrite this as $xa^3x^3a^{-1}xaxax^{-1}$.

However, different words may become equivalent due to the relations. For instance, the words $x$ and $x^{-1}$ represent the same group element because $x^2 = e$. This is an obvious equality, but some other equalities are not so obvious. For instance, $axa = x$ in the group -- this is not immediately clear, but it can be deduced from the relations:

$xax^{-1} = a^{-1} \implies axax^{-1} = e \implies axa = x$

Thus, the very large number (in fact, infinite number) of distinct words can correspond to a much smaller set of possible elements. We give a very simple argument to indicate that there are at most eight distinct elements:

$\! e, a, a^2, a^3, x, ax, a^2x, a^3x$

We start with any word. First note that since $x = x^{-1}$, all $x^{-1}$s can be rewritten as $x$.

We now claim that we can rewrite this word by bringing all the $a$s together and all the $x$s together. To do this, we need to bubble the $x$s past the $a$s. For this, we note the relations:

$xa = a^{-1}x$ (this can be deduced from $xax^{-1} = a^{-1}$)

and:

$xa^{-1} = ax$ (this can be deduced from $xax^{-1} = a^{-1}$ and using $x = x^{-1}$)

Thus, whenever we see an $a$ or $a^{-1}$ immediately right of an $x$, we can rewrite that part of the word pushing the $x$ to the right. Iterating this procedure often enough gets all powers of $a$ on the left, followed by all powers of $x$. We thus get a word of the form:

$a^mx^n$

Since $a^4 = x^2 = e$, $m$ can be reduced mod 4 and $n$ reduced mod 2, and we get the $4 \times 2 = 8$ possibilities listed above.

### Bounding from below

This is (in general) a lot trickier -- it means making sure that there is no further collapse than what we have already deduced from the relations. Roughly speaking, the only way to really do this is to actually construct a group -- with its full multiplication table or multiplication rule, and show that this rule is associative and satisfies all the relations. In practice, there are a number of theorems about presentations that we can use to short-circuit this process, but at this stage we don't have access to those theorems.

In our case, we have identified that there are at most eight distinct elements:

$\! e, a, a^2, a^3, x, ax, a^2x, a^3x$

In the process, we have made clever use of all the relations. However, how can we be sure that there isn't more scope to use the relations cleverly and cut the group down further, i.e., to find some way of showing that $x = a^2x$? This is the kind of question that haunts one when working with presentations. As mentioned, the basic tool is to actually construct the multiplication table once we're sure that we've used up all relations and verify that this multiplication table gives a group satisfying the relations.

## Constructing the actual multiplication table

### Abstract formulation

We first construct the multiplication table as an abstract set of rules on the set $\{ e, a, a^2, a^3, x, ax, a^2x, a^3x \}$:

• $e$ times any element is that element.
• $a^m$ times $a^n$ is $a^{m+n}$, where $m + n$ is reduced modulo 4.
• $a^m$ times $a^nx$ is $a^{m+n}x$, where $m +n$ is reduced modulo 4.
• $a^mx$ times $a^n$: This is the first nontrivial rule. The $x$ needs to be moved past $a$ a total of $n$ times. Each time, $xa$ becomes $a^{-1}x$, so we get $a^ma^{-n}x$ or $a^{m-n}x$, where $m - n$ is reduced mod 4.
• $a^mx$ times $a^nx$ is $a^{m - n}$ where $m - n$ is reduced modulo 4.

We can now verify computationally that:

• This rule is associative, for every triple of element types.
• $e$ is an identity element.
• Every element has an inverse: the inverse of $a^m$ is $a^{4-m}$, and the inverse of $a^mx$ is $a^mx$.
• The relations $a^4 = x^2 = e$ and $xax^{-1} = a^{-1}$ are all satisfied.

We thus obtain a group of order eight that realizes the presentation. Further, by the reasoning earlier, we also know that the group given by the presentation has order at most eight. We have thus bounded the group both from above and below, and obtained its precise description.

### Actual multiplication table

Working out the above abstract rules, we can construct the actual $8 \times 8$ multiplication table.

The row element is multiplied on the left and the column element is multiplied on the right.

Element $\! e$ $\! a$ $\! a^2$ $\! a^3$ $\! x$ $\! ax$ $\! a^2x$ $\! a^3x$
$\! e$ $\! e$ $\! a$ $\! a^2$ $\! a^3$ $\! x$ $\! ax$ $\! a^2x$ $\! a^3x$
$\! a$ $\! a$ $\! a^2$ $\! a^3$ $\! e$ $\! ax$ $\! a^2x$ $\! a^3x$ $\! x$
$\! a^2$ $\! a^2$ $\! a^3$ $\! e$ $\! a$ $\! a^2x$ $\! a^3x$ $\! x$ $\! ax$
$\! a^3$ $\! a^3$ $\! e$ $\! a$ $\! a^2$ $\! a^3x$ $\! x$ $\! ax$ $\! a^2x$
$\! x$ $\! x$ $\! a^3x$ $\! a^2x$ $\! ax$ $\! e$ $\! a^3$ $\! a^2$ $\! a$
$\! ax$ $\! ax$ $\! x$ $\! a^3x$ $\! a^2x$ $\! a$ $\! e$ $\! a^3$ $\! a^2$
$\! a^2x$ $\! a^2x$ $\! ax$ $\! x$ $\! a^3x$ $\! a^2$ $\! a$ $\! e$ $\! a^3$
$\! a^3x$ $\! a^3x$ $\! a^2x$ $\! ax$ $\! x$ $\! a^3$ $\! a^2$ $\! a$ $\! e$