A presentation of a group $G$ is an object that can be viewed in two ways:

a way of making $G$ as a quotient of the free group on some set $X$

a set $X$ of generators (from which we form the set $X^{- 1}$ of formal inverses to $X$ ^[1]), and a set $R$ of relators (which must be freely reduced words on $X \cup X^{- 1}$ ), such that every element of $G$ may be written as a product of the generators and such that by combining elements of $R$ we can obtain every possible way of expressing the identity element of $G$ as a product of elements of $X$ .

Every group $G$ has a presentation with $G$ as the set of generators, and the set of relators is the set containing every trivial word. Of course, this presentation is in general not unique: we may, for instance, add a new generator $t$ and the relator $t$ to any presentation to obtain an isomorphic presentation.

The above presentation corresponds to taking the quotient of the free group $F (G)$ on $G$ by the homomorphism $ϕ : F (G) \to G$ which sends a word $(a_{1}, a_{2}, \dots, a_{n})$ to the product $a_{1} a_{2} \dots a_{n}$ . This is an instance of the more widely-useful fact that every group is a quotient of a free group (proof).

Examples

The cyclic group $C_{2}$ on two elements has a presentation $⟨ x ∣ x^{2} ⟩$ . That is, it has just one generator, $x$ , and the relator $x^{2}$ tells us that $x^{2}$ is the identity $e$ . Notice that $⟨ x ∣ x^{4} ⟩$ would also satisfy the description that "there is one generator, and $x^{4}$ is the identity". However, the group corresponding to this presentation contains four elements, not two, so it is not $C_{2}$ . This demonstrates the fact that if we have a presentation $⟨ X ∣ R ⟩$ , and a group can be written in such a way that all the relators hold in the group, and the group can be generated by the elements of $X$ , that still doesn't mean the presentation describes the group; it could be that extra relations hold in the group that aren't listed in $R$ . (In this case, for example, $x^{2} = e$ is not listed in $⟨ x ∣ x^{4} ⟩$ .)

The presentation $⟨ x, y ∣ x y x^{- 1} y^{- 1} ⟩$ describes a group with two generators, such that the only nontrivial relation is $x y x^{- 1} y^{- 1} = e$ (and anything that can be built up from that). That relation may be written as $x y = y x$ : that is, $x$ and $y$ commute. This tells us that the group is abelian, since every generator commutes with every other generator. In fact, this group's elements are just words $x^{n} y^{m}$ for some integers $n, m$ ; this follows because, for instance, $x y x = x x y = x^{2} y$ , and in general we can pull all the instances of the letter $x$ (and $x^{- 1}$ ) out to the front of the word. Therefore we can write an element of this group as $(m, n)$ where $m, n$ are integers; hence the group is just $Z^{2}$ with pointwise addition as its operation.

The presentation $⟨ x, y ∣ x^{2}, y ⟩$ is just $C_{2}$ again. Indeed, we have a relator telling us that $y$ is equal to the identity, so we might as well just omit it from the generating set (because it doesn't add anything new to any word in which it appears).

The presentation $⟨ a, b ∣ a b a^{- 1} b^{- 2}, b a b^{- 1} a^{- 2} ⟩$ is a longwinded way to define the trivial group (the group with one element). To prove this, it is enough to show that each generator represents the identity, because then every word on the generators has been made up from the identity element so is itself the identity. We have access to the facts that $a b = b^{2} a$ and that $b a = a^{2} b$ in this group (because, for example, $a b a^{- 1} b^{- 2} = e$ ). The rest of the proof is an exercise.

Show solution

We have $a b = b^{2} a$ from the first relator; that is $b b a$ . But $b a = a^{2} b$ is the second relator, so that is $b a^{2} b$ ; hence $a b = b a^{2} b$ and so $a = b a^{2}$ by cancelling the rightmost $b$ . Then by cancelling the rightmost $a$ , we obtain $e = b a$ , and hence $a = b^{- 1}$ .

But now by the first relator, $a b = b^{2} a = b b a$ ; using that both $a b$ and $b a$ are the identity, this tells us that $e = b$ ; so $b$ is trivial.

Now $a = b^{- 1}$ and so $a$ is trivial too.

finite presentation/generation

direct products

semidirect products

^{^︎}
For example, if $X = a, b$ then $X^{- 1}$ is a set of new symbols which we may as well write $a^{- 1}, b^{- 1}$ .