Prove that a group of order 255 is a cyclicSolutionLet G be

Prove that a group of order 255 is a cyclic.

Solution

Let G be a group of order 255 = 3.5.17.

Prove that G is cyclic.

Proof:

Step 1: G has a unique subgroup of order 17 , say N which is normal in G.

Let n be the number of 17-Sylow subgroups of G. According to Sylow theory, n divides 255, so n is one of the following integers:

                                                                        1,3,5,17,15,51,85,255.

In addition, n should be 1 mod 17.

this forces n =1. So there is a unique subgroup N of order 17. N is normal in G because, as per Sylow, all p-Sylow subgroups are conjugate.

Step 2: N is contained in the center Z of G.

. As G acts on N by conjugation , we have a homomorphism

                                                                 f: G-> Aut (N) .

But N is isomorphic to Z₁₇ (group of prime order)

Now Aut (Z₁₅) is just the multiplicative ;group of units of Z₁₇, with order 16.

As 255/order (Ker f) divides 16 (by first isomorphism theorem) it follows that Ker f = G, (ie. the map f is trivial).

Thus N is contained in Z.

Step 3. If G is a group with N contained in Z (center) and G/N is cyclic , then G.is abelian.

This is a standard fact, proof goes like this. Let gN be a generator of G/N. If x is in G, then xN = (gN)^m for some integer m, hence xg^-m is in N, so x = g^m z for a central element z. So for x and y in G we have x = g^m z, y= gⁿ z, so xy = g^m+n z² , which is also equal to yx (note z is in the center).

So G is abelian.

Step 4: Applying it to our case , we obtain G is abelian (G/N has order 15 and any group of order 15 is cyclic)

Step 5: Any abelian group of order 255 must be isomorphic to Z₃ x Z₅ xZ₁₇, Hence cyclic