## Existence of Sylow subgroups

Let be a finite group such that where is a prime number, and . Then there exists a subgroup such that . (such a subgroup is called a Sylow subgroup).

**Proof:**

**1.** Let be the finite field with elements. For consider the space of invertible morphisms (or the space of matrices with elements in which are invertible). Then is a finite groupe with cardinal equal to

To see this note that every invertible matrix corresponds to a basis of . First we see that is finite because there are at most matrices with elements in . To find the cardinal of we need to count the bases of . We can choose the first element in ways because any element not equal to zero works.

Once we have chosen the first vector, the second cannot be a scalar multiple of the first (there are such scalars), therefore we are left with possibilities.

Once we have chosen vectors, the space of linear combinations of these vectors has elements, so we have choices at step . Multiply all these for and get the above formula.

We deduce that with . If we denote the set of matrices which have on the diagonal and zeros below the main diagonal, then is a subgroup of and has elements. Therefore is a -Sylow subgroup of .

**2. Lemma** Suppose is a group with and a subgroup of . Suppose has a -Sylow subgroup . Then there exists such that is a -Sylow subgroup of .

*Proof of the lemma:* acts on (left classes) by left translation, and the stabilizer of is . () acts also on by left translation and the above argument proves that the stabilizer of under this action is .

These stabilizers are all -groups since they are subgroups of . To finish it is enough to prove that for some the cardinal is prime to . But we know that is the cardinal of the orbit of under the action of and the sum of cardinals of orbits equals the cardinal of the set . Since is not divisible by ( is -Sylow), it follows that there exists such that is prime to . Therefore is a -Sylow subgroup of .

**3.** In general, if we know that is isomorphic to a subgroup of (Cayley’s theorem). Furthermore, can be regarded as a subgroup of by associating to each permutation the morphism defined on the canonical base by

In this way, we can see is isomorphic to a subgroup of . Apply points 1. and 2. and we are done.