## Sylow's Second Theorem

Sylow's Second Theorem

Let be a finite group of order n and let be a prime dividing then the number of distinct Sylow – subgroups (remember that if is the highest power of dividing then the Sylow – subgroup of is that subgroup which has order ). of is congruent to 1 (mod ).

Example: We can show that if is a group of order 91 then

1. has only one Sylow - 13 subgroup

2. must be cyclic. This is done as follows. The Sylow 13 – subgroups of are of order 13 and hence cyclic. Hence each Sylow 13 - subgroup is generated by a single element, and all such groups have trivial intersection (only the identity), since every element has order 13. If any two such subgroups have elements in common, those elements generate each group and the groups are identical. Each new 13 – Sylow subgroup contributes 12 new elements to all of order 13.

If has Sylow – 13 subgroups then m is congruent to 1 (mod 13) so =1,14,... but if then which is a contradiction since so – this proves the first part above.

Now we follow the the same reasoning for 7 - Sylow subgroups.

91=7*13 so the Slow 7 – subgroups must be cyclic of order 7. As for the Sylow 13 – subgroups, any two distinct subgroups must have trivial intersection. Suppose has l Sylow 7 – subgroups. Each such group contributes 6 distinct elements to of order 7. If we assume there are no elements order 91 the the class equation becomes 91=1+12+6l which requires l=13 but 13 is not congruent to 1 (mod 7), contradicting Sylow's Second Theorem. Thus has at least one element of order 91 and so is cyclic.

It is not true in general that if and are primes that any group of order is cyclic e.g. of order 6 is not cyclic, but it is true that if for prime with then  has exactly one Sylow – subgroup. Since it is the only such subgroup, it must be normal in 