# Finding the elements of a group given two generators and relations

While I'm at it let me just confirm a few results I got, cos if they are right then I think I got the hang of manipulating ordered products. So I got these relationships, could u tell me if they are right?
$aba = b$
$ab = ba^{4}$
$ba = a^{4}b$
$a^{2}b = aba^{4}$, which becomes, using the first equation, $a^{2}b=ba^{3}$

So my group has been reduced to 10 elements, which are $E, a, a^{2}, a^{3}, a^{4}, b, ab, ba, ba^{2}, ba^{3}$

NOTE - I have kept the elements of the form $ba^{k}$, but pasmith has used the other equality instead that $ba^{2} = a^{3}b$ etc.

lavinia
Gold Member
First of all this group has 10 elements. Your first list was correct.

In your second list you forgot that b$^{2}$ = id.

A group with two sets of generators of finite order,n amd m, does not have to be abelian to have order nxm.

Whenever the cyclic subgroup generated by one is normal this will be true. See if you can prove this.

In your example the cyclic subgroup of order 5 generated by b is normal. This is what the second relation says.

The relation aba$^{-1}$ = b$^{-1}$ defines a group no matter what the order of b. If b has order 4 for instance it is the dihedral group of order 8.

Good examples of easy finite groups can be made by starting with two orthogonal matricies with only 1,-1, and 0 entries and taking all of the multiples. You will find many interesting groups this way.

For instance the dihedral group of order 8 is generated by the matricies

a =

0 -1
1 0

which a a positive rotation of 90 degrees

and

a =

1 0
0 -1

which is a reflection around the y axis.

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