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.