# Thread: Order of the group

1. ## Order of the group

G is a group given by :
G = { q | ad - bc = 1} where a, b, c, d are integers modulo p, a prime number and

Find the o(G)

2. Originally Posted by poorna
G is a group given by :
G = { q | ad - bc = 1} where a, b, c, d are integers modulo p, a prime number and

Find the o(G)
let $\displaystyle L$ be the group of $\displaystyle 2 \times 2$ invertible matrices with entries in $\displaystyle \mathbb{F}_p.$ define the homomorphism $\displaystyle f: L \longrightarrow \mathbb{F}_p^{\times}$ by $\displaystyle f(A)=\det A.$ then $\displaystyle G=\ker f$ and $\displaystyle L/G \cong \mathbb{F}_p^{\times}.$ thus $\displaystyle |G|=\frac{|L|}{p-1}.$

to find $\displaystyle |L|$ look at the elements of $\displaystyle L$ as invertible transformations from $\displaystyle V$ to $\displaystyle V,$ where $\displaystyle V$ is a vector space of dimension $\displaystyle 2$ over $\displaystyle \mathbb{F}_p.$ let $\displaystyle \{v_1,v_2 \}$ be a basis for $\displaystyle V.$ let $\displaystyle T \in L.$

then $\displaystyle T(v_1)$ may be defined to be any non-zero element of $\displaystyle V.$ so $\displaystyle T(v_1)$ has $\displaystyle p^2 - 1$ options. now $\displaystyle T(v_2)$ can be anything in $\displaystyle V - <T(v_1)>.$ so there are $\displaystyle p^2-p$ options for $\displaystyle T(v_2).$

hence $\displaystyle |L|=(p^2-1)(p^2-p)$ and therefore $\displaystyle |G|=\frac{|L|}{p-1}=p(p^2-1). \ \ \Box$

the above method can be used to, in general, find the order of the group of all $\displaystyle n \times n$ matrices with determinant 1 and with entries from some finite field. (just copy the above proof!)

3. hey, this is a subgp. of GL2(Fp) whose order is p(p-1)(p^2 -1). Infact it is the normal subgp. Can this be used to solve the problem. We have been given this as the hint.

4. Originally Posted by poorna
hey, this is a subgp. of GL2(Fp) whose order is p(p-1)(p^2 -1). Infact it is the normal subgp. Can this be used to solve the problem. We have been given this as the hint.

No, no "the" normal subgroup but "a" normal subgroup, and what NCA explained you also explains why it is normal: it is the kernel of a homomorphism. I think the hint you're talking about is exactly the solution NCA gave you.

Tonio