Sylow's Second Theorem

Sylow's Second Theorem

Letbe a finite group of order n and letbe a prime dividingthen the number of distinct Sylow– subgroups (remember that ifis the highest power ofdividingthen the Sylow– subgroup ofis that subgroup which has order). ofis congruent to 1 (mod).

Example: We can show that ifis 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 ofare 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 toall of order 13.

IfhasSylow – 13 subgroups then m is congruent to 1 (mod 13) so=1,14,... but ifthen which is a contradiction sinceso– 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. Supposehas l Sylow 7 – subgroups. Each such group contributes 6 distinct elements toof 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. Thushas at least one element of order 91 and so is cyclic.

It is not true in general that ifandare primes that any group of orderis cyclic e.g.of order 6 is not cyclic, but it is true that ifforprime withthenhas exactly one Sylow– subgroup. Since it is the only such subgroup, it must be normal in

Add comment

Security code