0

Let $H$ be the subgroup of the group $G$ of all $2 \times 2$ non-singular matrices whose entries are integers modulo a given prime $p$ consisting of those and only those matrices in $G$ whose determinant is $1$.

What is the order of $H$? And how to find it?

I've already managed to find the order of $G$. It is $p^4 - p^3 - p^2 + p$.

Last but not least, I would also like to be able to compute the order of $G$ and that of $H$ in the general case corresponding to the $n \times n$ matrices for an arbitrary integer $n > 2$.

Saaqib Mahmood
  • 26,762
  • 1
    So... how did you find it? (Otherwise, what part of this do you need help with?) – Zev Chonoles Feb 16 '14 at 16:47
  • Zev Chonoles, please read my question carefully, and you'll see what I require help with. I require help with finding the total number of matrices of unit determinant with entries that are integers modulo a given prime $p$. – Saaqib Mahmood Feb 16 '14 at 16:59
  • mik, look carefully through my posting. You see, I've already figured out what you've suggested as a possible answer to my question. My question is a bit more involved. – Saaqib Mahmood Feb 16 '14 at 17:01
  • 1
    $G$ is denoted as $GL(2,p)$ and its order is correct, but you might want to write it as $(p^2-1)(p^2-p)$, which formula generalized more easily to different dimensions and fields ($GL(n,q)$, where $q$ is a prime-power). Your $H$ is denoted by $SL(2,p)$ and its order, as you figured out is $(p-1)p(p+1)$, which is easy to memorize if you write it like that! – Nicky Hekster Feb 16 '14 at 17:17

1 Answers1

2

Hint: $H$ is the kernel of $\det\colon G\to \mathbb F_p^\times$.

Hagen von Eitzen
  • 374,180
  • Hagen von Eitzen, thanks for your valuable hint. Since the set of all left (or right) cosets of $H$ is in one-to-one correspondence with $F_p^{*}$, we have order of $H = (p^4 - p^3 - p^2 + p) / (p-1) = p^3 - p$. Is it correct? If not, please supply the necessary correction. Also, I would like to come up with the answer through more elementary means, without making use of the homomorphisms. – Saaqib Mahmood Feb 16 '14 at 17:09