# A subgroup of $\mathrm{SL}_n(\mathbb{Z}/p\mathbb{Z})$

Let $$p$$ be an odd prime, and consider the group $$\{U\in \operatorname{SL}_n(\mathbb{Z}/p\mathbb{Z}) : U^{t}U=I \bmod p \}\subseteq \operatorname{SL}_n(\mathbb{Z}/p\mathbb{Z}).$$

I wonder what is the structure of this subgroup? Can the generators of this subgroup be written or calculated ?

We know that if $$U^{t}U=I$$ with $$U\in \operatorname{GL}_n(\mathbb{Z})$$, then $$U$$ must be a signed permutation. So I came up with the above subgroup.

• This is a perfectly reasonable question. The one line answer is to point to groupprops.subwiki.org/w/… . I'm trying to write up a longer answer now, although someone who knows group theory better could probably do a better job. Commented Mar 23, 2023 at 13:14

This is the special orthogonal group over the field $$\mathbb{Z}/p \mathbb{Z}$$. Or, rather, one of the "special orthogonal groups". For any invertible symmetric matrix $$Q$$, one can consider the group of matrices $$U$$ in $$\text{SL}_n$$ such that $$U Q U^T = Q$$, which forms a subgroup $$\text{SO}(Q)$$. If one allows matrices in $$\text{GL}_n$$ instead, this is $$O(Q)$$.

I'll limit myself to $$p$$ odd throughout.

"Split" versus "non-split" If $$n=2m+1$$ is odd, then any two nondegenerate quadratic forms define the same orthogonal group.

If $$n=2m$$ is even, then there are two classes of quadratic forms, called the "split" and the "non-split" class. The case you have described is split if $$p^{m} \equiv 1 \bmod 4$$ and is non-split if $$p^{m} \equiv 3 \bmod 4$$. I'll denote the split and non-split cases by $$SO_{2m}^+(p)$$ and $$SO_{2m}^-(p)$$.

Order of the group The order of $$SO_{2m+1}$$ is $$p^{m^2} \prod_{i=1}^m (p^{2i}-1).$$ The orders of $$SO_{2m}^+(p)$$ and $$SO_{2m}^-(p)$$ are $$p^{m(m-1)} (p^m-1) \prod_{i=1}^{m-1} (p^{2i}-1) \ \text{and} \ p^{m(m-1)} (p^m+1) \prod_{i=1}^{m-1} (p^{2i}-1)$$ respectively. Note, in particular, that they can't be isomorphic since they have different sizes.

Generators The orthogonal group $$O_n$$ is generated by reflections; for a vector $$x$$ with $$x \cdot x \neq 0$$, the reflection over $$x$$ is the linear map $$r_x(v) := v - \tfrac{x \cdot v}{x \cdot x} x$$. Reflections have determinant $$-1$$, so the special orthogonal group $$SO_n$$ is generated by pairs of reflections; the product $$r_x r_y$$ fixes the $$(n-2)$$-plane $$x^{\perp} \cap y^{\perp}$$, so we can also say that $$SO_n$$ is generated by rotations fixing planes of dimension $$n-2$$. A shorter list of generators can be found in Matrix Generators for the Orthogonal Groups, Rylands and Taylor , 1998.

Structure The orthogonal group $$O_n$$ has two characters to $$\{ \pm 1 \}$$: The determinant map, which sends every reflection to $$-1$$, and the spinor norm, which sends the reflection $$r_x$$ to $$\left( \tfrac{x \cdot x}{p} \right)$$ (this is the quadratic residue symbol). I'll write $$K_{2m+1}$$, $$K_{2m}^+$$ or $$K_{2m}^-$$ for the common kernel of these characters. So $$K$$ is an indexed $$2$$ subgroup of the corresponding $$SO$$.

For $$n=2m+1$$ odd, the group $$K_n$$ is the Chevalley group $$B_m(p)$$. The group $$B_1(p)$$ is isomorphic to $$\text{PSL}_2(p)$$, which is the alternating group $$A_4$$ if $$p =3$$ and is otherwise simple. I believe that $$B_m(p)$$ is simple for all $$m \geq 2$$ and $$p \geq 3$$, but I couldn't find a reference for this.

For $$n$$ even in the case you care about, where $$Q = \text{Id}_n$$, the group $$K_n$$ contains $$- \text{Id}_n$$, which generates the center. For the other quadratic form that you didn't use, the spinor norm of $$- \text{Id}_n$$ is $$-1$$, and the center of $$K_{n}$$ is trivial.

The quotients of $$K_{2m}^+(p)$$ and $$K_{2m}^-(p)$$ by their centers are the Chevalley group $$D_m(p)$$ and the twisted Chevalley group $${}^2 D_m(p^2)$$ respectively. We have $$D_2(p) \cong \text{PSL}_2(p) \times \text{PSL}_2(p)$$. Again, $$\text{PSL}_2(p)$$ is simple for any $$p \geq 5$$, whereas $$\text{PSL}_2(3) \cong A_4$$. I believe that $$D_m(p)$$ and $${}^2 D_m(p^2)$$ are simple for $$m \geq 3$$ (and $$p$$ odd), but again I couldn't find a reference

I'm using Wikipedia as my source for group theoretic notation.

• You wrote "The case you have described is split if $p=1$ mod $4$ and non-split if $p=3$ mod $4$." It seems to me that this is correct if $n=2$ mod $4$, but that if $4$ divides $n$, this form $\sum_{k=1}^n x_k^2$ is split regardless of the odd $p$.
– YCor
Commented Mar 23, 2023 at 14:41
• @YCor Thanks, you are right! That means groupprops.subwiki is wrong. I don't have an account there, do you? Commented Mar 23, 2023 at 14:44
• No, I don't either.
– YCor
Commented Mar 23, 2023 at 15:34