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Ben Webster ♦
Ben Webster ♦
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By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki group hasgroups have order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.

By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki group has order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.

By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki groups have order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.

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Ben Webster ♦
Ben Webster ♦
  • 44.7k
  • 12
  • 126
  • 260

By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki group has order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.