# Sylow Theorems

The **Sylow theorems** are a collection of results in the theory of finite groups. They give a partial converse to Lagrange's Theorem, and are one of the most important results in the field. They are named for P. Ludwig Sylow, who published their proof in 1872.

## The Theorems

Throughout this article, will be an arbitrary prime.

The three Sylow theorems are as follows:

**Theorem.**Every finite group contains a Sylow -subgroup.**Theorem.**In every finite group, the Sylow -subgroups are conjugates.**Theorem.**In every finite group, the number of Sylow -subgroups is equivalent to 1 (mod ).

As , the third theorem implies the first.

Before proving the third theorem, we show some preliminary results.

**Lemma 1.** Let and be nonnegative integers. Then

*Proof.* Let be a group of order (e.g., , and let be a set of size . Let act on the set by the law ; extend this action canonically to the subsets of of size . There are such subsets.

Evidently, a subset of is stable under this action if and only if . Thus the fixed points of the action are exactly the subsets of the form , for . Then there are fixed points. Therefore since the -group operates on a set of size with fixed points.

Let be a finite group, and let its order be , for some integer not divisible by .

**Lemma 2.** Let be the set of subsets of of size , and let act on by left translation. Suppose is an orbit of such that does not divide . Then has elements, and they are disjoint.

*Proof.* Since divides but is relatively prime to , it follows that divides ; in particular, . Since every element of is included in some element of ,
with equality only when the elements of are disjoint and when . Since equality does occur, both these conditions must be true.

**Theorem 3.** The number of finite subgroups of is equivalent to 1 (mod ).

*Proof.* Consider the action of on , as described in Lemma 2. Since
and every orbit of for which does not divide has elements, it follows that the number of such orbits is equivalent to 1 (mod ). It thus suffices to show that every such orbit contains exactly one Sylow -subgroup, and that every Sylow -subgroup is contained in exactly one such orbit.

To this end, let be an orbit of for which does not divide . Consider the equivalence relation on elements of , defined as " and are in the same element of ". Then is compatible with left translation by ; since the elements of are disjoint, is an equivalence relation. Thus the equivalence class of the identity is a subgroup of , which must have order . Since this is the only element of that contains the identity, it is the only group in .

Conversely, if is a Sylow -subgroup, then its orbit is its set of left cosets, which has size , which does not divide. Since orbits are disjoint, is contained in exactly one orbit of .

We now prove a more general theorem that implies the second Sylow theorem.

**Theorem 4.** Let be a Sylow -subgroup of , and let be a -subgroup of . Then is a subgroup of some conjugate of .

*Proof.* Consider the left operation of on , the set of left cosets of modulo . Since the order of is some power of , say , the size of each orbit must divide ; since there are cosets, and , it follows that some orbit must have size 1, i.e., there must be some such that is stable under the operation by . Then for all ,
i.e., stabilizes . Thus . It follows that , or .

**Corollary 5 (second Sylow theorem).** The Sylow -subgroups of are conjugates.

**Corollary 6.** Every subgroup of that is a -group is contained in a Sylow -subgroup of .

**Corollary 7.** Let be a Sylow -subgroup of , let be its normalizer, and let be a subgroup of that contains . Then is its own normalizer.

*Proof.* Let be an element of such that . Then is a Sylow -subgroup of . Since the Sylow -subgroups of are conjugates, there exists such that . It follows that normalizes , so . Hence . Thus the normalizer of is a subset of . Since the opposite is true in general, is its own normalizer.

**Corollary 8.** Let and be finite groups, and a homomorphism of into . Let be a Sylow -subgroup of . Then there exists a Sylow -subgroup of such that .

For is a -subgroup of ; this corollary then follows from Corollary 6.

**Corollary 9.** Let be as subgroup of and let be a Sylow -subgroup of . Then there exists a Sylow -subgroup of such that .

*Proof.* There must be some Sylow -subgroup of that contains . Since is a -subgroup of , it can be no larger than .

**Corollary 10.** Conversely, if is a Sylow -subgroup of and is a normal subgroup of , then is a Sylow -subgroup of .

*Proof.* Let be the greatest integer such that divides the order of . Let be the canonical homomorphism from to . Then is the kernel of the restriction of to . Since is isomorphic to , a -subgroup of of order not exceeding , it follows from Lagrange's Theorem that has order at least . Thus ; since is a -subgroup of , it follows that it is a Sylow -subgroup of .

Note that this is not true in general if is not normal. For instance is a Sylow 2-subgroup of the symmetric group and is a subgroup of , but their intersection, , is evidently not a Sylow 2-subgroup of .

**Corollary 11.** Let be a normal subgroup of , let be the canonical homomorphism. Then the image of every Sylow -subgroup of under is a Sylow -subgroup of ; furthermore, every Sylow -subgroup of is the image of a Sylow -subgroup of under .

*Proof.* Let be a Sylow -subgroup of ; let the order of be , where is a positive integer not divisible by . By Corollary 9, is a group of order ; it follows that is isomorphic to , a group of order , so is a Sylow -subgroup of . Now, let be a Sylow -subgroup of ; then there exists such that . Let be such that ; then .