Editing Sylow theorems (section)

=== Small groups are not simple ===
A more complex example involves the order of the smallest [[simple group]] that is not [[cyclic group|cyclic]]. [[Burnside's theorem|Burnside's ''p<sup>a</sup> q<sup>b</sup>'' theorem]] states that if the order of a group is the product of one or two [[prime power]]s, then it is [[solvable group|solvable]], and so the group is not simple, or is of prime order and is cyclic. This rules out every group up to order 30 {{nowrap|({{=}} 2 · 3 · 5)}}.

If ''G'' is simple, and |''G''| = 30, then ''n''<sub>3</sub> must divide 10 ( = 2 · 5), and ''n''<sub>3</sub> must equal 1 (mod 3). Therefore, ''n''<sub>3</sub> = 10, since neither 4 nor 7 divides 10, and if ''n''<sub>3</sub> = 1 then, as above, ''G'' would have a normal subgroup of order 3, and could not be simple. ''G'' then has 10 distinct cyclic subgroups of order 3, each of which has 2 elements of order 3 (plus the identity). This means ''G'' has at least 20 distinct elements of order 3.

As well, ''n''<sub>5</sub> = 6, since ''n''<sub>5</sub> must divide 6 ( = 2 · 3), and ''n''<sub>5</sub> must equal 1 (mod 5). So ''G'' also has 24 distinct elements of order 5. But the order of ''G'' is only 30, so a simple group of order 30 cannot exist.

Next, suppose |''G''| = 42 = 2 · 3 · 7. Here ''n''<sub>7</sub> must divide 6 ( =  2 · 3) and ''n''<sub>7</sub> must equal 1 (mod 7), so ''n''<sub>7</sub> = 1. So, as before, ''G'' can not be simple.

On the other hand, for |''G''| = 60 = 2<sup>2</sup> · 3 · 5, then ''n''<sub>3</sub> = 10 and ''n''<sub>5</sub> = 6 is perfectly possible. And in fact, the smallest simple non-cyclic group is ''A''<sub>5</sub>, the [[alternating group]] over 5 elements. It has order 60, and has 24 [[cyclic permutation]]s of order 5, and 20 of order 3.