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.

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    $\begingroup$ 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. $\endgroup$ Commented Mar 23, 2023 at 13:14

1 Answer 1


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.

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    $\begingroup$ 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 $p$. $\endgroup$
    – YCor
    Commented Mar 23, 2023 at 14:41
  • $\begingroup$ @YCor Thanks, you are right! That means groupprops.subwiki is wrong. I don't have an account there, do you? $\endgroup$ Commented Mar 23, 2023 at 14:44
  • $\begingroup$ No, I don't either. $\endgroup$
    – YCor
    Commented Mar 23, 2023 at 15:34

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