There are at most two finite simple groups of any order

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Statement

Let n be a natural number. Then, there are at most two finite simple groups of order n.

The smallest value of n for which there are two non-isomorphic simple groups of order n is 20160 = 2^6 \cdot 3^2 \cdot 5 \cdot 7. The two groups in this case are the alternating group of degree eight (which is also isomorphic to the projective special linear group PSL(4,2)) and projective special linear group:PSL(3,4).

Description of relevant pairs

Infinite families

Note that there are examples other than the infinite families given below.

Below, q = p^r is a prime power denoting the size of the field over which we are considering stuff.

Condition on q Condition on n First family (Chevalley notation) First family (description) Second family (Chevalley notation) Second family (description) Order
odd (when q is even, the corresponding groups are in fact isomorphic) n > 2 (for n = 1 or n = 2, the corresponding groups are in fact isomorphic) B_n(q) Chevalley group of type B, more explicitly this is \Omega_{2n+1}(q), the intersection of kernels of the Dickson invariant and spinor norm in the orthogonal group O(2n+1,q) C_n(q) projective symplectic group PSp(2n,q) \! q^{n^2} \left[\prod_{i=1}^n (q^{2i} - 1)\right]/\operatorname{gcd}(2,q-1)

First few examples

These include both the infinite families and the examples not arising from infinite families.

Order First group Second group
20160 alternating group:A8 (isomorphic to PSL(4,2)) projective special linear group:PSL(3,4)
4585351680 Chevalley group of type B:B3(3) projective symplectic group:PSp(6,3)

Proof

All known proofs of this fact employ the classification of finite simple groups, along with some explicit computations.