Group presentation

by Patrick Stevens Jul 22 2016 updated Jul 27 2016

Presentations are a fairly compact way of expressing groups.

[summary: A presentation $~$\langle X \mid R \rangle$~$ of a group is, informally, a way of specifying the group by a set $~$X$~$ of generators together with a set $~$R$~$ of relators. Every element of the group is some product of generators, and the relators tell us when a product is trivial.]

[summary(Technical): A presentation $~$\langle X \mid R \rangle$~$ of a group $~$G$~$ is a set $~$X$~$ of generators and a set $~$R$~$ of relators which are words on $~$X \cup X^{-1}$~$, such that $~$G \cong F(X) / \llangle R \rrangle^{F(X)}$~$ the [-normal_closure] of $~$\llangle R \rrangle$~$ with respect to the Free group $~$F(X)$~$. ]

A presentation $~$\langle X \mid R \rangle$~$ of a group $~$G$~$ is an object that can be viewed in two ways:

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 $~$\phi: 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).


%%hidden(Show solution): We have $~$ab = b^2 a$~$ from the first relator; that is $~$b ba$~$. But $~$ba = a^2 b$~$ is the second relator, so that is $~$b a^2 b$~$; hence $~$ab = b a^2 b$~$ and so $~$a = b a^2$~$ by cancelling the rightmost $~$b$~$. Then by cancelling the rightmost $~$a$~$, we obtain $~$e = ba$~$, and hence $~$a = b^{-1}$~$.

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

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

[todo: finite presentation/generation] [todo: direct products] [todo: semidirect products]