Difference between revisions of "Sylow Theorems"
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Latest revision as of 12:25, 9 April 2019
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.
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 .