Finite special linear group SL_n(q)

Let $q=p^f$ , with $p$ prime and $f\geq1$ an integer, and choose the finite field $\mathbb F_q$ . Let $n\geq1$ be an integer. The special linear group $\mathrm{SL}_{n}(q)$ is the group of determinant-one $n\times n$ matrices over $\mathbb F_q$ , with matrix multiplication.

Order

The group order is $\frac{1}{q-1}\prod_{j=0}^{n-1}(q^n-q^j)$ . The determinant map from the general linear group onto $\mathbb F_q^\times$ has this group as kernel. Dividing the general linear group order by $q-1$ gives the formula.

Simplicity conditions

In size one the group is trivial. For size $m\geq2$ , the center is $\{\lambda I:\lambda^m=1\}$ , of order $\gcd(m,q-1)$ . Its quotient is $\mathrm{PSL}_m(q)$ , which is simple except at $(m,q)=(2,2),(2,3)$ . Thus the special linear group itself is simple exactly when its center is trivial and these two exceptions are absent.

Field and category convention

The parameter $q$ must be a prime power. This is a finite abstract group; the field and its vector space supply a construction, rather than a vector-space structure on the matrix group itself.

Determinant sequence

The inclusion into the general linear group and the determinant give an exact sequence

1\longrightarrow\mathrm{SL}_{n}(q)\longrightarrow\mathrm{GL}_{n}(q)\xrightarrow{\det}\mathrm{GL}_1(q)\longrightarrow1.

The diagonal matrix with entries $a,1,\ldots,1$ maps to any prescribed $a\in\mathbb F_q^\times$ , proving surjectivity. In size one, this is the identity map on $\mathbb F_q^\times$ , with trivial kernel.

References

Peter J. Cameron, Notes on Classical Groups, §2.1, Theorems 2.2–2.3, pp. 15–16: orders, determinant and scalar kernels; §2.4, Theorem 2.10, pp. 20–22: simplicity.