# Thread: prime field and order

1. ## 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?

2. ## 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?

3. ## 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.

4. ## 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?)

5. ## 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

6. ## 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).