Hi,
I have come across the following result in a book:
GL_{n}(F_{p}) ⊇ F^{*}_{pn}
Does anyone know if this has a name or even know how I can prove it? Thanks in advance
there is an isomorphic copy of F^{*}_{pn} in GL_{n}(F_{p}).
the latter group can be regarded as the set of invertible nxn matrices with entries in F_{p}. the former group is the group of units for F_{pn}, which can be regarded as an n-dimensional vector space over F_{p}.
if α is in F^{*}_{pn}, then the map x→αx is an invertible linear map (F_{p})^{n}→(F_{p})^{n}, so it has a representation as an element of GL_{n}(F_{p}).
a concrete example where p = 2 and n = 2:
then F_{4} = {0,1,α,α+1}, where α is a root of x^{2}+x+1 in F_{2}[x]. using {1,α} as a basis, we can write this as:
F_{4} = {(0,0),(1,0),(0,1),(1,1)}. our basis for F_{4} is {(1,0),(0,1)}.
consider the mapping x→αx. note that this sends 1→α and sends α→α^{2}. now since α^{2}+α+1 = 0, α^{2} = -α-1 = α+1 (since 1 = -1 in F_{2}).
so in the basis {(1,0),(0,1)}, this mapping has the matrix:
[0 1]
[1 1].
the identity element 1 in F_{4}, clearly yields the identity 2x2 matrix. the only other non-zero element of F_{4} is α+1.
the map x→(α+1)x sends (1,0) to (1,1), and (0,1) to (1,0), yielding the matrix:
[1 1]
[1 0],
giving us a 3-element subgroup of GL_{2}(F_{2}) (which has 6 elements).