# What is the cardinality of $\mathrm{SL}_2(\mathbb{Z}_p)$, where $p$ is a prime? [duplicate]

I was solving past exam papers and stuck on the following problem:

Let $$p$$ be a prime number .Suppose $$G$$ be the group of all $$2 \times 2$$ matrix over $$\Bbb Z_p$$,with determinant $$1$$ under matrix multiplication. Then order of $$G$$ is which of the following :

1. $$p(p-1)(p+1)$$

2. $$p^2(p-1)$$

3. $$p^3$$

4. $$p^2(p-1)+p$$

Can someone explain it? Thanks and regards to all.

• The group of $n\times n$ matrices over a field $F$ that have determinant $1$ is known as the special linear group of degree $n$ over $F$ and is denoted $\mathrm{SL}_n(F)$ or $\mathrm{SL}(n,F)$. I think this question has already been asked here, though I can't find a duplicate at the moment; in the meantime, the reasoning used in this analogous question about the general linear group may help you (or, now that you know the name of this group, you could easily look up the answer on Google). – Zev Chonoles Jul 9 '13 at 4:14
• Do you mean $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ $\not = \mathbb{Z}_p$? The set $\mathbb{Z}/p\mathbb{Z}$ of integers mod $p$ has $p$ elements, whereas the set $\mathbb{Z}_p$ of $p$-adic integers has many many more! – Mike May 4 '19 at 19:47

Can you calculate $|{\rm GL}_2({\bf F}_p)|$? Invertible linear maps correspond to invertible $2\times2$ matrices, or equivalently ordered bases of ${\bf F}_p^2$: there are $\square$ many vectors to choose from for the first vector in a basis, and $\square$ many vectors to choose from for the second vector, so there are $\square$-many bases total.

Now consider the kernel of $\det:{\rm GL}_2({\bf F}_p)\to{\bf F}_p^\times$ in light of the $1$st isomorphism theorem.

Here's a way. Take $$GL(2,Z_p)$$ and consider the matrix row vectors as $$r_1,r_2$$. Clearly $$r_1$$ has $$p^2$$ choices(why?). But as we want $$\det \ne 0$$ $$r_1$$ has $$p^2-1$$ choices. Now $$r_2$$ has $$p^2-p$$ choices(as the rows must be linearly independent so $$r_2 \ne cr_1$$ for some $$c \in Z_p$$). So $$\vert GL_2(Z_p) \rvert = (p^2-1)(p^2-p)$$ Now consider the homomorphism $$f : GL \rightarrow Z_p$$ given by $$f(A)=\det(A)$$.

So, by 1st iso theorem, $$\lvert SL_2(Z_p) \rvert=(p^2-1)(p^2-p)/(p-1)$$.

If $\mathbb{F}_{q}$ denotes finite field of order q then order of $\operatorname{SL}(n,\mathbb{F}_{q})$is $\frac{1}{q-1}\prod_{{i=0}}^{{n -1}}(q^{n}-q^{i})$
• Yes, it is. $\phantom{ }$ – Martin Brandenburg Jul 9 '13 at 6:19