Sylow subgroups exist

Statement
Let $$G$$ be a fact about::finite group and $$p$$ be a prime number. Then, there exists a $$p$$-fact about::Sylow subgroup $$P$$ of $$G$$: a subgroup whose order is a power of $$p$$ and whose index is relatively prime to $$p$$.

Note that when $$p$$ does not divide the order of $$G$$, the $$p$$-Sylow subgroup is trivial, so the statement gives interesting information only when $$p$$ divides the order of $$G$$. This statement is often viewed as a part of a more general statement called Sylow's theorem.

Other parts of Sylow's theorem

 * Sylow implies order-dominating: Given any $$p$$-Sylow subgroup and any $$p$$-subgroup, the $$p$$-Sylow subgroup contains a conjugate of the $$p$$-subgroup.
 * Sylow implies order-conjugate: All $$p$$-Sylow subgroups are conjugate. This follows directly from the previous part.
 * Congruence condition on Sylow numbers: The number of $$p$$-Sylow subgroups is congruent to $$1$$ modulo $$p$$.

Stronger forms of existence

 * Every Sylow subgroup intersects the center nontrivially or is contained in a centralizer: Every Sylow subgroup of a group either has a nontrivial intersection with the center of the group, or it is contained in the centralizer of some non-central element.

Analogues for Hall subgroups
The analogous statement does not hold for Hall subgroups. A Hall subgroup is a subgroup whose order and index are relatively prime. A $$\pi$$-Hall subgroup is a Hall subgroup such that the set of primes dividing its order is contained in $$\pi$$, and the set of primes dividing its index is disjoint from $$\pi$$.


 * Hall subgroups need not exist: Given a set $$\pi$$ of primes, there may not exist a $$\pi$$-Hall subgroup.
 * Hall subgroups exist in finite solvable: If, however, the finite group is solvable, then it has $$\pi$$-Hall subgroups for all prime sets $$\pi$$.
 * Hall's theorem: This states that a finite group is solvable if and only if it has $$\pi$$-Hall subgroups for every prime set $$\pi$$.

Analogues in other algebraic structures

 * Sylow subloops exist for Sylow primes in finite Moufang loop
 * 2-Sylow subloops exist in finite Moufang loop
 * 3-Sylow subloops exist in finite Moufang loop
 * Hall subloops exist in finite solvable Moufang loop

Facts used

 * 1) uses::Lucas' theorem (more specifically, uses::Lucas' theorem prime power case)
 * 2) uses::Fundamental theorem of group actions: There is a bijection between the coset space of the stabilizer of an element and the orbit of that element. In particular, the size of the orbit equals the index of the stabilizer.
 * 3) uses::Lagrange's theorem
 * 4) uses::Class equation of a group
 * 5) uses::Cauchy's theorem for abelian groups
 * 6) uses::Central implies normal: Any subgroup inside the center is normal.

Proof by action on subsets
Given: A finite group $$G$$ of order $$n = p^rm$$ where $$p$$ is prime, $$r$$ is a nonnegative integer, and $$p$$ does not divide $$m$$.

To prove: $$G$$ has a subgroup of order $$p^r$$.

Proof: Here are some observations regarding this action:

Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory by doing the above proof in reverse taking a cyclic group (although a purely algebraic proof also exists). ,

Proof by conjugation action
Given: A finite group $$G$$ of order $$n = p^rm$$, where $$p$$ is prime, $$r$$ is a nonnegative integer, and $$p$$ does not divide $$m$$.

To prove: $$G$$ has a subgroup of order $$p^r$$.

Proof: We prove the claim by induction on the order of $$G$$. Specifically, we assume that the result is true for all groups of order smaller than the order of $$G$$.

For the case $$r = 0$$, the proof is direct since the trivial group is a subgroup of order $$p^r$$. Thus, we assume $$r > 0$$. In particular, $$p$$ divides the order of $$G$$.

Consider the class equation of $$G$$ (fact (4)):

$$|G| = |Z(G)| + \sum_{i=1}^r |G:C_G(g_i)|$$

where $$c_1,c_2,\dots,c_r$$ are the conjugacy classes of non-central elements and $$g_i$$ is an element of $$c_i$$ for each $$i$$.

We consider two cases:

Case that $$p$$ divides the order of $$Z(G)$$ :

Case that $$p$$ does not divide the order of $$Z(G)$$