The free group on k generators is the group Fk = ⟨a1,...,ak⟩. See generators and relations for groups for precise definition and explanation of what "generated" means here (or see the short explanation below). There are no relations for this group. One can also define free groups on infinite sets of generators in the obvious manner; however, note that products of these generators are still finite (we have no topology, hence no convergence).
Elements of the group have a canonical form ai1n1...aimnm, where 1≤ij≤k is the index of some generator, 0≠nj∈Z is a nonzero integer, and ij≠ij+1.
Any element can be translated to canonical form by combining adjacently equal ai's. For instance,
a2 a1 a32 a3-4 a12 =
a2 a1 a3-2 a12.
Thus, the
word problem on the free group is (easily)
solvable by
computer program.
Making groups from the free group
Any group with k generators is a quotient group of Fk. The relations of the group are a normal subgroup N, and the group is simply Fk/N.
Obviously, Fj is a subgroup (as well as a quotient
group) of Fk whenever j≤k. Somewhat amazingly, the
reverse also holds! We shall show that Fk is a subgroup
of F2 -- thus, study of this one group encompasses all
of (finite, and even countable) group theory!
To generate a copy Fk≅G≤F2=⟨a,b⟩, we look
at the subgroup generated by the elements
⟨e1=a1ba-1,...,ek=akba-k⟩.
Then we have that e
im=a
ib
ma
-i,
so we can readily identify powers of our generators. And
multiplication
behaves nicely: if i≠j, then
eimejn = aibmaj-ibna-j.
Thus, we can easily "
decode" any finite
product of e
i's when phrased as a product of a's and b's. The rigourous argument to show
our G is free proceeds in this fashion.
The free group can also be found as a quotient in numerous places. Perhaps the
most famous is the identification of a subgroup of PSL2(Z) (itself a quotient group of the group SL2(Z) of integer 2×2 matrices with determinant 1) as the free group F2. This identification is what makes hyperbolic geometry interesting.