By Mahdi Alosh
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4. Unlabelled graphs and trees Usually, it is easier to count labelled objects than unlabelled. We saw that there are 2n(n−1)/2 labelled graphs, and nn−2 labelled trees, on n vertices. Any unlabelled object can be labelled in at most n! ways, with equality if it has no non-trivial automorphisms. For graphs, it is well-known that almost all graphs have no non-trivial automorphisms, and so the number of unlabelled graphs on n vertices is asymptotic to 2n(n−1)/2 /n! Reﬁning the asymptotics involves considering symmetry.
The subgroup of index 2 consisting of matrices of determinant 1 is the special orthogonal group. The subgroup of scalars has order 2. The resulting projective special orthogonal group is NOT simple in general. There is (usually) a further subgroup of index 2, which is not so easy to describe. The spinor norm In general, orthogonal groups are generated by reﬂections: rv : x → x − 2 B(x, v) v. B(v, v) The reﬂections have determinant −1, so the special orthogonal group is generated by even products of reﬂections.
The stabilizer of the point (1, 0, 0, . . , 0) consists (modulo scalars) of λ 0 matrices . It has a normal Abelian subgroup consisting of matrices vM 1 0 . v In−1 These matrices encode elementary row operations, and it is an elementary theorem of linear algebra that every matrix of determinant 1 is a product of such matrices. To prove that P SLn (q) is perfect, it suﬃces to prove that these matrices (transvections) are commutators. If n ≥ 3, observe that 100 100 1 00 1 1 0, 0 1 0 = 0 1 0.