Let
F be any field. Prove that
SLn(F) is a subgroup of
GLn(F).
Proof. Let
F be any field. Recall that the general linear group is defined as:
GLn(F)={A ∣ A is an n×n invertible matrix with entries from F}
and the special linear group is defined as:
SLn(F)={A∈GLn(F) ∣ det(A)=1}.
We want to show that if
A,B∈SLn(F), then
A−1∈SLn(F) and
AB∈SLn(F).
Assume for the first case that
A∈SLn(F). Since
det(A)=1, and therefore inverse, we have that:
det(A−1)=det(A)1=11=1.
So
A−1∈SLn(F).
For the second case, assume that
A,B∈SLn(F). Then
det(A)=det(B)=1. Now we have that:
det(AB)=det(A)det(B)=1.
Therefore,
AB∈SLn(F).
So
SLn(F) is a subgroup of
GLn(F).