 Finite symplectic group order

ID: finite-symplectic-group-order

Counting successive symplectic pairs gives

∣ Sp_{2 m} (q) ∣ = q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1) .

For each nonzero first vector there are

q^{2 m - 1}

partners with pairing one, after which its nondegenerate plane splits off. The exact defining-characteristic part is

q^{m^{2}}

, realized by an upper symplectic unitriangular subgroup.

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