# Group presentation

A pre­sen­ta­tion $$\langle X \mid R \rangle$$ 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 $$X$$noteFor ex­am­ple, if $$X = \{ a, b \}$$ then $$X^{-1}$$ is a set of new sym­bols which we may as well write $$\{ a^{-1}, b^{-1} \}$$. ), 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 $$\phi: F(G) \to G$$ 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).

# Examples

• The cyclic group $$C_2$$ on two el­e­ments has a pre­sen­ta­tion $$\langle x \mid x^2 \rangle$$. That is, it has just one gen­er­a­tor, $$x$$, and the re­la­tor $$x^2$$ tells us that $$x^2$$ is the iden­tity $$e$$. No­tice that $$\langle x \mid x^4 \rangle$$ 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 $$\langle X \mid R \rangle$$, 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 $$\langle x \mid x^4 \rangle$$.)

• The pre­sen­ta­tion $$\langle x, y \mid xyx^{-1}y^{-1} \rangle$$ de­scribes a group with two gen­er­a­tors, such that the only non­triv­ial re­la­tion is $$xyx^{-1}y^{-1} = e$$ (and any­thing that can be built up from that). That re­la­tion may be writ­ten as $$xy=yx$$: 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, $$xyx = xxy = x^2y$$, 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 $$\mathbb{Z}^2$$ with poin­t­wise ad­di­tion as its op­er­a­tion.

• The pre­sen­ta­tion $$\langle x, y \mid x^2, y \rangle$$ 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 $$\langle a, b \mid aba^{-1}b^{-2}, bab^{-1}a^{-2} \rangle$$ 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 $$ab = b^2 a$$ and that $$ba = a^2 b$$ in this group (be­cause, for ex­am­ple, $$aba^{-1} b^{-2} = e$$). The rest of the proof is an ex­er­cise.

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. <div><div>

finite pre­sen­ta­tion/​gen­er­a­tion di­rect prod­ucts semidi­rect products

Parents:

• Group

The alge­braic struc­ture that cap­tures sym­me­try, re­la­tion­ships be­tween trans­for­ma­tions, and part of what mul­ti­pli­ca­tion and ad­di­tion have in com­mon.