Let be a finite group. Let be the group of permutations on Define the map by for all It is easy to see that is a group homomorphism:

for all Thus i.e. is a group homomorphism.

**Lemma 1**. If the order of is then is a product of disjoint cycles of length

*Proof*. It is pretty obvious: choose Then, since we will have

So

is a cycle of length in the decomposition of Now let Then is another cycle of length in the decomposition of Clearly and are disjoint. We can continue this process until no element in is left.

**Problem 1**. Let be a group with where is odd. Suppose that has a cyclic Sylow -subgroup. Prove that has a (normal) subgroup of index

**Solution.** Define the map by for all Here is considered as the multiplicative group of order 2 and is the signature of the permutation of Clearly is a group homomorphism because it is the composite of two group homomorphisms. If is the identity element of then obviously Now, if is a cyclic subgroup of of order then has order and thus, by Lemma 1, where each is a cycle of length Thus and hence because is an odd integer. So we have proved that is onto. Let Then and therefore

**Lemma 2.** Let be a finite group and let be a normal subgroup of Let be a prime divisor of and suppose that is a Sylow -subgroup of Then is a Sylow -subgroup of

*Proof*. Let and where Since is a subgroup of both and we have for some So in order to show that is a Sylow -subgroup of we only need to show that Now,

But is a subgroup of because is normal. Thus and so

We will continue this discussion in the next post.