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Math Help - prime field and order

  1. #1
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    prime field and order

    Hi, I need some help with this question.

    Let F=F_p be a field and V=F_p^n.
    a. The numbers of bases of V for n=2 is equal to the order of GL_2 (F)
    b. Prove the order of GL_2 (F) is p(p+1)(p-1)^2.
    c. Prove the order of SL_2 (F) is p(p+1)(p-1).

    Working:
    If v_1,...,v_n is a basis of V then there is an invertible matrix with these vectors as columns.
    Note that v_1 can be any non-zero vector, then v_2 can be any vector in V that is not on the line spanned by v_1, and same for v_3 to v_n. So there are p^n-1 choices for v_1, p^n-p choices for v_2, p^n-p^2 choices for v_3, etc.

    So the order of GL_n (F_p) is \prod_{i=0}^{n-1} (p^n-p^i)

    So this gives me the answer to part b, but how do I work out part a and c?

    Thanks for any help you provide.
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  2. #2
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    Re: prime field and order

    a) you need a bijection from the set of all bases for V to GL2(F). you already have mapping (you found one in part (b)), so you just have to show that this mapping is a bijection.

    c) suppose A is in GL2(F). how many possible values can det(A) have? can you partition GL2(F) into cosets using this information?
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  3. #3
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    Re: prime field and order

    a) define a map \phi : Gl_2(F_p) \to B where B is the set of all bases of V.

    So we choose a basis of V, e_1,...,e_n and then for T \in GL_2(F_p) we define \phi(T) = Te_1,...,Te_n. And since T is non-singular, \phi(T) \in B.

    But for some stupid reason at the moment I am coming up blank from where to go from here to show that there is a bijection between V and GL2(F_p).


    c) So SL_2(F_p) has 2x2 matrices with determinant 1. So I can define a map

    det: GL_2(F_p) \to F_p^{\times} which sends A \mapsto det(A)

    And using
    det(A \cdot B)=det(A) \cdot det(B)

    and

    A= \begin{bmatrix}a & 0 \\0 & 1 \end{bmatrix} \mapsto det(A)=a

    Now this is a surjective group homomorphism with ker(det)=SL_2(F_p)

    Thus |GL_2(F_p) : SL_2(F_p)|=|im(det)|=p-1

    Thus, |SL_2(F_p)|=\frac{|GL_2(F_p)|}{p-1}=p(p+1)(p-1)

    Does this make sense?

    Is there a better way to do this part?

    Thanks for ur response.
    Last edited by shelford; October 1st 2011 at 07:16 PM.
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  4. #4
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    Re: prime field and order

    part (c) looks good.

    for (a) you want to show φ is bijective. is φ onto? that is, given a basis {v1,v2,..,vn} in B, can we find (at least one) invertible linear map T with T(ej) = vj?

    put another way, can the map that sends ej-->vj be extended to a linear map on V? why is this (extended) map necessarily invertible (and thus in GL2(Fp))?

    to show φ is injective...suppose φ(A) = φ(B). then A(ej) = B(ej), for each basis vector ej, so...

    (technically, for this to work, you should use "ordered bases" for V. why?)
    Last edited by Deveno; October 1st 2011 at 09:53 PM.
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  5. #5
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    Re: prime field and order

    Thanks, but for a I am still having trouble finding at invertible linear map T such that T(ej)=vj.
    And I don't get why 'ordered bases' is a must.

    Thanks
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  6. #6
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    Re: prime field and order

    look we have a "chosen" basis {e1,e2,...,en} for V, and if {v1,v2,...vn} is any element of B, we can define a bijection from {e1, e2,...,en} to {v1,v2,...,vn},

    let's call this bijection f. define a linear map T:V-->V by setting T(a1e1+a2e2+....+anen) = a1f(e1)+a2f(e2)+....+anf(en)

    (can you see that this map is automatically linear?)

    note that T(ej) = T(0e1+0e2+...+1ej+...+0en) = 0f(e1)+0f(e2)+...+1f(ej)+...+0f(en) = f(ej) = vj.

    and T^-1 is the map that sends vj-->ej, that is:

    T^-1(a1v1+a2v2+...+anvn) = a1e1+a2e1+...+an, so T has an inverse, so it is in GL2(Fp).

    so φ(T) = {T(e1),...,T(en)} = {v1,...vn}, thus φ is onto.

    now suppose φ(A) = φ(B), for A,B in GL2(Fp). this means {A(e1),...,A(en)} = {B(e1),...,B(en)}.

    and here is where using "ordered bases" is helpful: if we are using ordered bases, we have that A(ej) = B(ej) for every j.

    hence A = B, so φ is injective (if we don't use ordered bases, then we could have φ(A) = {v2,v1,...,vn} = {v1,v2,...,vn} = φ(B)

    which complicates things. ultimately, we're still just dealing with linear combinations of the basis elements, it just makes the statement of what we are

    trying to prove somewhat messier).
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