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 and repeated use of $x = x^{- 1}$ :

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

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 $3 \times 2 = 8$ 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:

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

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.