Editing Sylow theorems (section)

==Example applications==
Since Sylow's theorem ensures the existence of p-subgroups of a finite group, it's worthwhile to study groups of prime power order more closely. Most of the examples use Sylow's theorem to prove that a group of a particular order is not [[Simple group|simple]]. For groups of small order, the congruence condition of Sylow's theorem is often sufficient to force the existence of a [[normal subgroup]]. 
;Example-1: Groups of order ''pq'', ''p'' and ''q'' primes with ''p''&nbsp;<&nbsp;''q''. 
;Example-2: Group of order 30, groups of order 20, groups of order ''p''<sup>2</sup>''q'', ''p'' and ''q'' distinct primes are some of the applications. 
;Example-3: (Groups of order 60): If the order |''G''|&nbsp;=&nbsp;60 and ''G'' has more than one Sylow 5-subgroup, then ''G'' is simple.

=== Cyclic group orders ===
Some non-prime numbers ''n'' are such that every group of order ''n'' is cyclic.  One can show that ''n'' = 15 is such a number using the Sylow theorems:  Let ''G'' be a group of order 15 = 3 · 5 and ''n''<sub>3</sub> be the number of Sylow 3-subgroups. Then ''n''<sub>3</sub> <math>\mid</math> 5 and ''n''<sub>3</sub> ≡ 1 (mod 3). The only value satisfying these constraints is 1; therefore, there is only one subgroup of order 3, and it must be [[normal subgroup|normal]] (since it has no distinct conjugates). Similarly, ''n''<sub>5</sub> must divide 3, and ''n''<sub>5</sub> must equal 1 (mod 5); thus it must also have a single normal subgroup of order 5. Since 3 and 5 are [[coprime]], the intersection of these two subgroups is trivial, and so ''G'' must be the [[internal direct product]] of groups of order 3 and 5, that is the [[cyclic group]] of order 15. Thus, there is only one group of order 15 ([[up to]] isomorphism).

=== Small groups are not simple ===
A more complex example involves the order of the smallest [[simple group]] that is not [[cyclic group|cyclic]]. [[Burnside's theorem|Burnside's ''p<sup>a</sup> q<sup>b</sup>'' theorem]] states that if the order of a group is the product of one or two [[prime power]]s, then it is [[solvable group|solvable]], and so the group is not simple, or is of prime order and is cyclic. This rules out every group up to order 30 {{nowrap|({{=}} 2 · 3 · 5)}}.

If ''G'' is simple, and |''G''| = 30, then ''n''<sub>3</sub> must divide 10 ( = 2 · 5), and ''n''<sub>3</sub> must equal 1 (mod 3). Therefore, ''n''<sub>3</sub> = 10, since neither 4 nor 7 divides 10, and if ''n''<sub>3</sub> = 1 then, as above, ''G'' would have a normal subgroup of order 3, and could not be simple. ''G'' then has 10 distinct cyclic subgroups of order 3, each of which has 2 elements of order 3 (plus the identity). This means ''G'' has at least 20 distinct elements of order 3.

As well, ''n''<sub>5</sub> = 6, since ''n''<sub>5</sub> must divide 6 ( = 2 · 3), and ''n''<sub>5</sub> must equal 1 (mod 5). So ''G'' also has 24 distinct elements of order 5. But the order of ''G'' is only 30, so a simple group of order 30 cannot exist.

Next, suppose |''G''| = 42 = 2 · 3 · 7. Here ''n''<sub>7</sub> must divide 6 ( =  2 · 3) and ''n''<sub>7</sub> must equal 1 (mod 7), so ''n''<sub>7</sub> = 1. So, as before, ''G'' can not be simple.

On the other hand, for |''G''| = 60 = 2<sup>2</sup> · 3 · 5, then ''n''<sub>3</sub> = 10 and ''n''<sub>5</sub> = 6 is perfectly possible. And in fact, the smallest simple non-cyclic group is ''A''<sub>5</sub>, the [[alternating group]] over 5 elements. It has order 60, and has 24 [[cyclic permutation]]s of order 5, and 20 of order 3.

=== Wilson's theorem ===
Part of [[Wilson's theorem]] states that

:<math>(p-1)! \equiv -1 \pmod p</math>

for every prime ''p''. One may easily prove this theorem by Sylow's third theorem. Indeed,  
observe that the number ''n<sub>p</sub>'' of Sylow's ''p''-subgroups  
in the symmetric group ''S<sub>p</sub>'' is {{sfrac|1|''p''&nbsp;&minus;&nbsp;1}} times the number of p-cycles in ''S<sub>p</sub>'', ie. {{math|(''p''&nbsp;&minus;&nbsp;2)!}}. On the other hand, {{math|''n''<sub>''p''</sub> ≡ 1 (mod&nbsp;''p'')}}. Hence, {{math|(''p''&nbsp;&minus;&nbsp;2)! ≡ 1 (mod&nbsp;''p'')}}. So, {{math|(''p''&nbsp;&minus;&nbsp;1)! ≡ &minus;1 (mod&nbsp;''p'')}}.

=== Fusion results ===
[[Frattini's argument]] shows that a Sylow subgroup of a normal subgroup provides a factorization of a finite group.  A slight generalization known as '''Burnside's fusion theorem''' states that if ''G'' is a finite group with Sylow ''p''-subgroup ''P'' and two subsets ''A'' and ''B'' normalized by ''P'', then ''A'' and ''B'' are ''G''-conjugate if and only if they are ''N<sub>G</sub>''(''P'')-conjugate.  The proof is a simple application of Sylow's theorem: If ''B''=''A<sup>g</sup>'', then the normalizer of ''B'' contains not only ''P'' but also ''P<sup>g</sup>'' (since ''P<sup>g</sup>'' is contained in the normalizer of ''A<sup>g</sup>'').  By Sylow's theorem ''P'' and ''P<sup>g</sup>'' are conjugate not only in ''G'', but in the normalizer of ''B''.  Hence ''gh''<sup>−1</sup> normalizes ''P'' for some ''h'' that normalizes ''B'', and then ''A''<sup>''gh''<sup>−1</sup></sup> = ''B''<sup>h<sup>−1</sup></sup> = ''B'', so that ''A'' and ''B'' are ''N<sub>G</sub>''(''P'')-conjugate.  Burnside's fusion theorem can be used to give a more powerful factorization called a [[semidirect product]]: if ''G'' is a finite group whose Sylow ''p''-subgroup ''P'' is contained in the center of its normalizer, then ''G'' has a normal subgroup ''K'' of order coprime to ''P'', ''G'' = ''PK'' and ''P''∩''K'' = {1}, that is, ''G'' is [[p-nilpotent group|''p''-nilpotent]].

Less trivial applications of the Sylow theorems include the [[focal subgroup theorem]], which studies the control a Sylow ''p''-subgroup of the [[derived subgroup]] has on the structure of the entire group.  This control is exploited at several stages of the [[classification of finite simple groups]], and for instance defines the case divisions used in the [[Alperin–Brauer–Gorenstein theorem]] classifying finite [[simple group]]s whose Sylow 2-subgroup is a [[quasi-dihedral group]].  These rely on [[J. L. Alperin]]'s strengthening of the conjugacy portion of Sylow's theorem to control what sorts of elements are used in the conjugation.