How to do I solve the number of cases for which $ad - bc$ is equal to one and $ad$, $bc$ are not equal to $0$?

Question

# Problem 26 *

(a) $\ $ Let $G$ be the group of all $2 \times 2$ matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a$, $b$, $c$, $d$ are integers modulo $p$, $p$ a prime number, such that $ad-bc \neq 0$. $G$ forms a group relative to matrix multiplication. What is $o(G)$?

(b) $\ $ Let $H$ be the subgroup of the $G$ of part (a) defined by $$ H = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : ad-bc = 1 \right\}. $$ What is $o(H)$?

Please verify my solution to part (b). To solve the problem I took different cases -

EDIT - Some people are asking to me how I arrived at each of these cases -

1. $ad = 0$. It restrict $bc = −1$. $(2p − 1)(p − 1)$ is the total number of cases. EDIT - The number ways of choosing $a$, $d$ for which $ad = 0$ is: $2p − 1$. On the other hand, the number of ways of choosing $b$, $c$ for which $bc = −1$ is: $p − 1$. Thus when $ad = 0$, we can choose $a$, $b$, $c$, $d$ in $(2p − 1)(p − 1)$.

2. Similar number of cases are for $ad = 1$ and $bc = 0$. EDIT - Analogous to previous case, we get number of ways of choosing $a$, $b$, $c$, $d$ as $(2p − 1)(p − 1)$.

3. Number of cases for which $ad$ and $bc$ are not equal to zero. I solved it and I got $((p-1)(p-1) - (p-1))(p-1)$. EDIT - $(p-1)(p-1)$ is the number of ways in which we can choose $ad$ not equal to zero and we subtract $(p-1)$ so that we eliminate the cases in which $ad = 1$ and then I multiply by $(p-1)$ (number of cases in which we can choose $b$).

The problem is from Topics in Algebra by Herstein.

Answer 1

EDIT: you got it right I just messed up seeing that your expression and mine coincided. Apologies.

You got your first two cases correct I believe, giving you 2(2p-1)(p-1). But I think you strayed with your last case.

First, choose $ad$. It can't be $1$ and it can't be $0$, or we're back with your previous cases, so you have $p-2$ choices for $ad$. Then, choose $a$ in any of $p-1$ ways, and then $d$ is fixed. Now, $bc$ is fixed by $ad$, $b$ can be chosen in any of $p-1$ ways, and $c$ is fixed by $b$ and $bc$, so we have

$$(\text{choices for }\ ad)(\text{choices for }\ a)(\text{choices for }\ b) = (p-2)(p-1)(p-1).$$

Add that to what you've already got and you should be good to go.

nb I see elements with order p, p-1 and 2 in your group, so be sure your sum is divisible by those.

Answer 2

To begin with, the column vectors must be linearly independent. So the first column can be any of $p^2-1$ non-zero vectors, the second any of $p^2-p$ vectors not in the span of the first. This gives us $(p^2-1)(p^2-p)$ matrices with non-zero determinant. As we see from scaling the first column by a non-zero constant, each non-zero value of determinant occurs with the same frequency. We conclude that the determinant is $1$ in $$\frac{(p^2-1)(p^2-p)}{p-1}=p^3-p$$ cases. If your counting leads to the same final result, it is correct.

Answer 3

Your counting is accurate.

But, for a proof I would also like to see the computation for the following results that you have stated or used:

1. The number of ways of choosing $b$ and $c$ for which $bc = -1$ is $p-1$ (used in the first case).
2. The number of ways of choosing $a$ and $d$ for which $ad = 1$ is $p-1$ (used in the third case).
3. The number of ways of choosing $b$ and $c$ for which $bc = 1 - ad$, for fixed $a$ and $d$ such that $ad \neq 1$, is $p-1$ (also used in the third case).

Of course, all these are special cases of one fact, namely that the number of pairs of integers $(x,y)$ such that $0 \leq x,y < p$ and $xy = c \pmod p$, where $c$ is a non-zero integer, is $p - 1$.

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Additionally, there are superior ways to calculate $o(H)$. @HagenvonEitzen gives my favourite method, because it generalises very easily to $n \times n$ matrices. On the other hand, the case by case analysis becomes nightmarish, even for $n = 3$.

Answer 4

In $\mathbb Z_p$

\begin{array}{lcl} \text{there are} & p-1 &\text{ways to make $1+bc = 0$} \\ \text{there are} & 2p-1 &\text{ways to make $1+bc = 1$} \\ \text{there are} & p-1 &\text{ways to make $1+bc = 2$} \\ \text{there are} & p-1 &\text{ways to make $1+bc = 3$} \\ \text{there are} & p-1 &\text{ways to make $1+bc = p-1$} \\ &\vdots \\ \text{there are} & p-1 &\text{ways to make $1+bc = p-1$} \\ \hline \text{there are} & 2p-1 &\text{ways to make $ad = 0$} \\ \text{there are} & p-1 &\text{ways to make $ad = 1$} \\ \text{there are} & p-1 &\text{ways to make $ad = 2$} \\ \text{there are} & p-1 &\text{ways to make $ad = 3$} \\ &\vdots \\ \text{there are} & p-1 &\text{ways to make $ad = p-1$} \\ \end{array}

Number of ways to make $ad-bc=1$ are

$$(p-1)(2p-1) + (2p-1)(p-1) +(p-2)(p-1)^2 = p^3-p$$