Put
For every nonzero integer ,
Let
These are disjoint nonempty subsets of . If , then
so . Similarly, if , then
so .
The ping-pong lemma now identifies the subgroup generated by and with
Both generators have infinite order, so this is a free group of rank two inside the special linear group .
Use the injective group homomorphism from part (a). If , its image is an integral matrix . Choose a prime number that does not divide one nonzero entry of . The reduction modulo a prime in an integral matrix group homomorphism
then sends to a nonidentity element. Its target is a finite group, so the composite map separates from the identity. Hence is a residually finite group.

Articles by others on the same topic (0)

There are currently no matching articles.