- Sylow subgroups exist: For any finite group and any prime , has a subgroup whose order is the largest power of dividing . Such a subgroup is termed a -Sylow subgroup.
Proof using existence of Sylow subgroups
Given: A finite group , a prime divisor of the order of .
To prove: has a subgroup of order exactly .
Proof: By fact (1), has a -Sylow subgroup, say . Since divides the order of , is nontrivial. Pick a non-identity element . The order of divides the order of , and is hence a power of , say , where . Thus, the element is an element of order , and the cyclic subgroup it generates thus has order exactly .
Proof without using existence of Sylow subgroups
The idea is to consider the action of by cyclic permutations on the set:
First, note that the action does indeed send the set to itself: cyclic permutation of a word corresponds to a conjugation operation on the product, which preserves the identity. Explicitly:
Hence, if the original word gives the identity element, so does the new word.
The set has size . Using the class equation of a group action (or otherwise), note that the orbits all have size 1 or , so the number of orbits of size 1 (i.e., the number of fixed points) is congruent to mod . The latter number is a multiple of , hence the number of fixed points is a multiple of . Note that the point with all coordinates the identity element is a fixed point, so there is at least one fixed point. Thus, there must be at least fixed points. Any fixed point other than the "all-identity fixed point" is of the form where is the identity element, and we thus get an element of order .