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

Solution:

Let n17 be the number of Sylow 17 subgroup of G: We know n17 divides 15 and is congruent
to 1 mod 17. Thus n17 = 1 and there is a unique Sylow 17 subgroup, call it P: Now G/P is a group of
order 15. If we look at the number of Sylow 5 subgroups of G/P we see this number must divide 3 and
be congruent to 1 mod 5. Thus there is just one such subgroup Q\' and it must be normal. By the fourth
isomorphism theorem we know there is a subgroup Q in G such that Q/P = Q\' in G/P: This subgroup
will be normal and of order 85. (You could alternately argue that if P\' was any Sylow 5 subgroup of
G then PP\' is a subgroup of G of order 85 and since its index is the smallest prime dividing the order
of G we can conclude it is normal.) Clearly P<= Q is normal and since 5 + 1; 10 + 1 and 15 + 1 does
not divide 17 the Sylow 5 subgroup R of Q is normal in Q: So Q = PR; P intersect R = {1} and P and Q are
both normal in Q. Thus Q = P *Q and since P and Q have prime orders we know they are cyclic.
Thus Q = (Z/17Z) * (Z/5Z) = (Z/85Z) is a cyclic subgroup of G: Now if n3 is the number of Sylow
3 subgroups then n3 divides 85 and is congruent to 1 mod 3. The only way for this to happen is for
n3 = 1 (since 4, 7, 10, 13, 16 and 19 do not divide 85). Thus the Sylow 3 subgroup S of G is normal in
G: Since G = QS;Q intersect S = {1}and Q and S are normal subgroups of G we see G = Q * S = Z/255Z: (proved)