Problem 3 30 points Use Sylows theorem to prove that every g

Problem 3 (30 points). Use Sylow’s theorem to prove that every group of order 255 is isomorphic to Z/255Z.

Solution

Suppose G is a finite group of order |G|=255=3517. We will show that G must be cyclic.

there must be only one Sylow subgroup P of order 17. Thus PP must be a normal subgroup. Since 17 is prime the subgroup P is cyclic, say P=a. It is not difficult to see that every group of order 15 is cyclic, so G/P=bP is also cyclic. Now P(bP)^|b|=P, so |bP|=15|b|.

If |b|=255| we are done. Therefore we can assume |b|=15|. We will show that ab=ba which implies that ab has order 255 since 15 and 17 are coprime. Because P is normal, bab1=a^i for some integer ii. Using this fact we get b^2ab^2=ba^ib^1=(bab^1)^i=a^i^2 and by induction b^kab^k=a^i^k for all k1. Thus a=b^15ab^15=a^i^15, which gives i^151mod17. By Fermat\'s little theorem i^161mod17 and since 15 and 16 are coprime, i1mod17. Therefore bab^1=a, proving the claim.

This exercise is a special case of a more general fact. Using the idea of this solution and the fact that in a group of squarefree order the Sylow subgroup corresponding to the largest prime is normal, you can prove that every group of order nn is cyclic when nn and (n) are coprime ( is the totient function).