Constructing dihedral group:D8 from its presentation

This is a survey article intended for expository and illustrative purposes. The style in this article is thus more verbose and pedagogical than most pages. Links are provided to other, more concise pages.

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:

$⟨ a, x ∣ a^{4} = x^{2} = e, x a x^{- 1} = a^{- 1} ⟩$

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

$⟨ a, x ∣ a^{4} = x^{2} = e, x a x = a^{- 1} ⟩$

And yet another is:

$⟨ a, x ∣ a^{4} = x^{2} = x a x a = e ⟩$

For advanced users

Advanced users, who don't want to wade through a prolonged explanation, can simply read through the following more advanced stuff:

Presentation of semidirect product is disjoint union of presentations plus action by conjugation relations: Applying this to cyclic group:Z4 and cyclic group:Z2 acting by the inverse map should be enough to convince you that this is a group of order eight given as a semidirect product.
Presentation of dihedral groups
Read about the Todd-Coxeter algorithm for a general procedure to construct groups using their presentations.

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 $x a a a x x x a^{- 1} x a x a x^{- 1}$ . Using the shorthand for words, we can rewrite this as $x a^{3} x^{3} a^{- 1} x a x a x^{- 1}$ .

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, $a x a = x$ in the group -- this is not immediately clear, but it can be deduced from the relations:

$x a x^{- 1} = a^{- 1} ⟹ a x a x^{- 1} = e ⟹ a x a = x$

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:

$e, a, a^{2}, a^{3}, x, a x, a^{2} x, a^{3} x$

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:

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

and:

$x a^{- 1} = a x$ (this can be deduced from $x a x^{- 1} = a^{- 1}$ 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:

$a^{m} x^{n}$

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.

If this is unclear for you, here's an example: [SHOW MORE]

Consider the word:

$a x a^{3} x a^{7} x^{- 1} a^{- 1} x a^{2} x^{2}$

First, note that we can reduce all exponents of $a$ mod 4 and exponents of $x$ mod 2, and we get:

$a x a^{- 1} x a^{- 1} x a^{- 1} x a^{2}$

The second and third letter need to be interchanged. We rewrite $x a^{- 1} = a x$ to get:

$a a x x a^{- 1} x a^{- 1} x a^{2}$

Note that we now have two successive $x$ s, which we can collapse and remove as $x^{2} = e$ , and we get:

$a a a^{- 1} x a^{- 1} x a^{2}$

We can collapse the powers of $a$ on the left and get:

$a x a^{- 1} x a^{2}$

We need to interchange the second and third letter again, and use $x a^{- 1} = a x$ again, to get:

$a^{2} x x a^{2}$

We now use that $x^{2} = e$ and thus get:

$a^{2} a^{2}$

This becomes

a^{4}

, which reduces to

e

.

Here's another example:[SHOW MORE]

$x^{3} a^{6} x^{2} a x a^{- 1} a^{- 1} x^{- 1}$

We first simplify all the exponents of $a$ , mod 4 and all the exponents of $x$ mod 2, and get:

$x a^{2} a x a^{2} x$

Further simplify to:

$x a^{- 1} x a^{2} x$

We rewrite the left most

x a^{- 1}

as

a x

and get

a x^{2} a^{2} x

. Remove the

x^{2}

now and get

a^{3} x = a^{- 1} x

which is the final answer.

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:

$e, a, a^{2}, a^{3}, x, a x, a^{2} x, a^{3} x$

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 $x = a^{2} x$ ? 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 ${e, a, a^{2}, a^{3}, x, a x, a^{2} x, a^{3} x}$ :

$e$ 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 $a^{n} x$ is $a^{m + n} x$ , where $m + n$ is reduced modulo 4.
$a^{m} x$ times $a^{n}$ : This is the first nontrivial rule. The $x$ needs to be moved past $a$ a total of $n$ times. Each time, $x a$ becomes $a^{- 1} x$ , so we get $a^{m} a^{- n} x$ or $a^{m - n} x$ , where $m - n$ is reduced mod 4.
$a^{m} x$ times $a^{n} x$ 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 $a^{m} x$ is $a^{m} x$ .
The relations $a^{4} = x^{2} = e$ and $x a x^{- 1} = a^{- 1}$ 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	$e$	$a$	$a^{2}$	$a^{3}$	$x$	$a x$	$a^{2} x$	$a^{3} x$
$e$	$e$	$a$	$a^{2}$	$a^{3}$	$x$	$a x$	$a^{2} x$	$a^{3} x$
$a$	$a$	$a^{2}$	$a^{3}$	$e$	$a x$	$a^{2} x$	$a^{3} x$	$x$
$a^{2}$	$a^{2}$	$a^{3}$	$e$	$a$	$a^{2} x$	$a^{3} x$	$x$	$a x$
$a^{3}$	$a^{3}$	$e$	$a$	$a^{2}$	$a^{3} x$	$x$	$a x$	$a^{2} x$
$x$	$x$	$a^{3} x$	$a^{2} x$	$a x$	$e$	$a^{3}$	$a^{2}$	$a$
$a x$	$a x$	$x$	$a^{3} x$	$a^{2} x$	$a$	$e$	$a^{3}$	$a^{2}$
$a^{2} x$	$a^{2} x$	$a x$	$x$	$a^{3} x$	$a^{2}$	$a$	$e$	$a^{3}$
$a^{3} x$	$a^{3} x$	$a^{2} x$	$a x$	$x$	$a^{3}$	$a^{2}$	$a$	$e$