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

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- June 2nd 2012, 05:24 AMElectricMiscellaneous Theorem in Linear Algebra
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 - June 5th 2012, 12:03 AMDevenoRe: Miscellaneous Theorem in Linear Algebra
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).