# Group Presentations - do they determine the group

1. Jan 23, 2012

### Math Amateur

I am seeking to gain a good understanding of group presentations

Currently I have the following general question:

"Does a group presentation completely determine a particular group?"

The textbooks I am reading seem to indicate that a group presentation does actually determine/specify the group.

For example on page 31 of James and Liebeck: Representations and Characters of Groups we find:

"Let G be the dihedral group $D_{2n} = D_8 = <a,b: a^4 = b^2 = 1, b^{-1}ab = a^{-1}>$

Is this a complete specification of the dihedral group - i.e. does this presentation completely determine or specify the dihedral group $D_8$?

Surely it does not - because we additionally need to know that (or do we?)

a = (1 2 3 4) [ rotation of a sqare clockwise through the origin - see attached]

and

b = (2 4) [reflection about the line of symmetry through vertex 1 and the origin - see attached]

Possibly we also need to know that the elements of the group are

$D_8 = \{ 1, a, a^2, a^3, b, ba, ba^2, ba^3 \}$

but I suspect this can be deduced from the given relations.

#### Attached Files:

• ###### Dihedral Group D8 - elements a and b.pdf
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Last edited: Jan 24, 2012
2. Jan 24, 2012

### morphism

Yes, certainly. A presentation <S|R> with generators S and relations R is just shorthand for the quotient of the free group F(S) on S by the normal subgroup generated by R - in particular, it's a group. So the equality G=<S|R> is an equality of groups (or, maybe more precisely, an isomorphism of groups).

3. Jan 24, 2012

### Math Amateur

So, just to re-confirm this - a presentation completely specifies a group?

We do not, in the case of the group $D_8$ even need to know the nature of the genarators a and b?

4. Jan 24, 2012

### lavinia

[/itex]

The presentation defines the relations between the generators of the group. This determines the group up to isomorphism.

If the group,G, has n generators, then there is a homomorphism from the free group on n generators onto G that takes each generator in the free group to the corresponding generator in G. The kernel of this homomorphism is the complete set of relations in G.

If H is isomorphic to G then the group of relations is the same.

Take your example of the dihedral group of order 8.
It is the group generated by a rotation of the plane by 90 degrees and a reflection about the y axis.

Now look at the group of linear transformation of R^4 generated by the two matrices

0 0 0 1
1 0 0 0
0 1 0 0 A
0 0 1 0

0 0 1 0
0 1 0 0
1 0 0 0 B
0 0 0 1

This group is isomorphic to the dihedral group of order 8.

This is because A$^{4}$ = B$^{2}$ = Id and BAB = A$^{3}$

As you can see, even though this is a different group it has the same presentation as the first group.

Last edited: Jan 24, 2012
5. Jan 24, 2012

### Math Amateur

Hi Lavinia

Thanks for the help

Much appreciated

Peter