# Presentation of a group

(Redirected from Finitely presented group)

In mathematics, one method of defining a group is by a presentation. One specifies a set S of generators so that every element of the group can be written as a product of some of these generators, and a set R of relations among those generators. We then say G has presentation

[itex]\langle S \mid R\rangle.[itex]

Informally, G has the above presentation if it is the "freest group" generated by S subject only to the relations R. Formally, the group G is said to have the above presentation if it is isomorphic to the quotient of a free group on S by the normal subgroup generated by the relations R.

As a simple example, the cyclic group of order n has the presentation

[itex]\langle a \mid a^n = 1\rangle.[itex]

Every group has a presentation, and in fact many; a presentation is often the most compact way of describing the structure of the group.

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## Background

A free group on a set S = {si} is a group where each element can be uniquely described as a finite length product of the form:

x1a1 x2a2 ... xnan

where each xi is a member of S, and xixi+1 for any i; and each ai is any non-zero integer.

If G is any group, and S is a generating subset of G, then every element of G is also of the above form; but in general, these products will not uniquely describe an element of G.

For example, the dihedral group D8 can be generated by a rotation, r, of order 4; and a flip, f, of order 2; and certainly any element of D8 is a product of r 's and f 's.

However, we have, for example, r f r = f, r3 = r−1, etc.; so such products are not unique in D8. Each such product equivalence can be expressed as an equality to the identity; such as

r f r f = 1
r 4 = 1
f 2 = 1

Informally, we can consider these products on the left hand side as being elements of the free group F = <r,f>, and can consider the subgroup R of F which is generated by these strings; each of which would also be equivalent to 1 when considered as products in D8.

If we then let N be the subgroup of F generated by all conjugates x−1 R x of R, then N is a normal subgroup of F; and each element of K, when considered as a product in D8, will also evaluate to 1. Thus D8 is isomorphic to the quotient group F/N. We then say that D8 has presentation

[itex]\langle r, f \mid r^4 = f^2 = (rf)^2 = 1\rangle.[itex]

## Formal definition

Let S be a set and let <S> be the free group on S; that is the set of all words on S. Let R be a set of words on S, so R is a subset of <S>. To form a group with presentation <S|R> the idea is take the largest quotient of <S> so that each element of R gets identified with the identity element. Note that R may not be a subgroup, let alone a normal subgroup of <S>, so we cannot take a naive quotient by R. The solution is to take the normal closure N of R in <S>, which is defined as the smallest normal subgroup in <S> which contains R. The group <S|R> is then defined as the quotient group

[itex]\langle S \mid R \rangle = \langle S \rangle / N.[itex]

The elements of S are called the generators of <S|R> and the elements of R are called the relations. A group G is said to have the presentation <S|R> if it is isomorphic to <S|R>.

It is a common practice write relations in the form x = y where x and y are words on S. What this means is that y−1xR. This has the intuitive meaning that the images of x and y are supposed to be equal in the quotient group.

A group is said to have a finite presentation if it has a presentation for which the set of generators [itex]S[itex] and relations [itex]R[itex] are both finite.

## Examples

The following table lists some examples of presentations for commonly studied groups. Note that in each case there are many other presentations that are possible. The presentation listed is not necessarily the most efficient one possible.

the free group on S [itex]\langle S \mid \varnothing \rangle[itex] A free group is "free" in the sense that it is subject to no relations.
Cn, the cyclic group of order n [itex]\langle a \mid a^n \rangle[itex]
D2n, the dihedral group of order 2n [itex]\langle r, f \mid r^n, f^2, (rf)^2 \rangle[itex] Here r represents a rotation and f a reflection
D, the infinite dihedral group [itex]\langle r, f \mid f^2, (rf)^2 \rangle[itex]
Dicn, the dicyclic group [itex]\langle r, f \mid r^{2n} = 1, r^n = f^2, frf^{-1} = r^{-1} \rangle[itex] The quaternion group is a special case when n = 2
Z × Z [itex]\langle x, y \mid xy=yx \rangle[itex]
Zm × Zn [itex]\langle x, y \mid x^m, y^n, xy=yx \rangle[itex]
the free abelian group on S [itex]\langle S \mid R \rangle[itex] where R is the set of all commutators of elements of S
the symmetric group, Sn generators: [itex]\sigma_1, \ldots, \sigma_{n-1}[itex]
relations:
• [itex]{\sigma_i}^2 = 1[itex],
• [itex]\sigma_i\sigma_j = \sigma_j\sigma_i \mbox{ if } j \neq i\pm 1[itex],
• [itex]\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}[itex]

The last set of relations can be transformed into

• [itex]{(\sigma_i\sigma_{i+1}})^3=1[itex]

using [itex] {\sigma_i}^2=1 [itex].

Here [itex]\sigma_i[itex] is the permutation that swaps the ith element with the i+1 one. The product [itex]\sigma_i \sigma_{i+1}[itex] is a 3-cycle on the set [itex] \{i,i+1,i+2\} [itex].
the braid group, Bn generators: [itex]\sigma_1, \ldots, \sigma_{n-1}[itex]

relations:

• [itex]\sigma_i\sigma_j = \sigma_j\sigma_i \mbox{ if } j \neq i\pm 1[itex],
• [itex]\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}[itex]
Note the similarity with the symmetric group; the only difference is the removal of the relation [itex]{\sigma_i}^2 = 1[itex].
the tetrahedral group, TA4 [itex]\langle s,t \mid s^2, t^3, (st)^3 \rangle[itex]
the octahedral group, OS4 [itex]\langle s,t \mid s^2, t^3, (st)^4 \rangle[itex]
the icosahedral group, IA5 [itex]\langle s,t \mid s^2, t^3, (st)^5 \rangle[itex]

## Some theorems

Every group G has a presentation. To see this consider the free group <G> on G. Since G clearly generates itself one should be able to obtain it by a quotient of <G>. Indeed, by the universal property of free groups there exists a unique group homomorphism φ : <G> → G which covers the identity map. Let K be the kernel of this homomorphism. Then G clearly has the presentation <G|K>. Note that this presentation is highly inefficient as both G and K are much larger than necessary.

Every finite group has a finite presentation.

The negative solution to the word problem for groups states that there is no general algorithm which, for a given presentation <S|R> and two words u, v, decides whether u and v describe the same element in the group.

## References

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy