**Problem 2**. Let be a group with where is odd. Suppose that has a cyclic Sylow -subgroup. Prove that has a normal subgroup of order

**Solution**. Let be a cyclic Sylow -subgroup. We solve the problem using induction over the exponent of in There is nothing to prove if If then we are done by Problem 1. Now suppose By Problem 1, has a normal subgroup of index i.e. By Lemma 2, is a Sylow -subgroup of Clearly is cyclic because it is a subgroup of the cyclic group Thus, by the induction hypothesis, has a subgroup such that is normal in and So we only need to show that is normal in We will do this in two steps:

*Step 1*. is the unique subgroup of order in To see this, suppose is another subgroup of order in Then is a subgroup of because is normal in Thus for some integer But and so which gives us because is odd. But and so Therefore

*Step 2. * is normal in To see this, let and put Then, since is normal in we have So and are both subgroups of and Hence by step 1.

**Corollary 1**. Non-abelian groups of order are not simple.

*Proof*. Let be a non-abelian group of order Then a Sylow -subgroup of has order and so it is cyclic. So, by Problem 2, has a normal subgroup of order

We can extend the result in Problem 2 to a much stronger one:

**Problem 3**. Let be a group with where is the smallest prime divisor of and Suppose that has a cyclic Sylow -subgroup. Prove that has a normal subgroup of order

**Solution**. Let be a cyclic Sylow -subgroup. Let and be the centralizer and the normalizer of in respectively. By the lemma in this post, is isomorphic to a subgroup of Since and is cyclic, we have

Thus But since and is the smallest prime divisor of we have

Thus must divide So if then and hence, since we will have which is absurd. Therefore and the result now follows from the Burnside’s Normal Complement Theorem.

**Corollary 2**. If is a non-abelian group and where is the smallest prime divisor of and then is not simple.

*Proof*. Every Sylow -subgroup of has order and so it is cyclic. Thus, by Problem 3, has a normal subgroup of order and so is not simple.

**Corollary 3**. Non-abelian groups of square-free order are not simple.