Sylow subgroups exist

Statement

Let $G$ be a finite group and $p$ be a prime number. Then, there exists a $p$-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.

Related facts

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 on solvability: This states that a finite group is solvable if and only if it has $\pi$-Hall subgroups for every prime set $\pi$.

Proof

Proof by action on subsets

Let $n$ be the order of $G$. Let $p^r$ be the higher power of $p$ that divides the order of $G$. Clearly, a subgroup of $G$ is a $p$-Sylow subgroup if and only if it has order $p^r$. If we denote $n/p^r$ by $m$, then the index of any $p$-Sylow subgroup must be $m$.

Consider the action of $G$ by left multiplication on the set of subsets of $G$ of size $p^r$. Here are some observations regarding this action:

• The size of the set on which $G$ acts is the number of subsets of size $p^r$ in $G$. This is a binomial coefficient, and an appeal to Lucas' theorem reveals that its value is relatively prime to $p$.
• Since the total cardinality of the set is relatively prime to $p$, there must exist at least one orbit under the action of $G$ which has size relatively prime to $p$. Let $S$ be the isotropy subgroup for a point in this orbit.
• Now, since the size of the orbit is relatively prime to $p$, the index of $S$ is relatively prime to $p$, and hence a divisor of $m$.
• On the other hand, given any subset $T$ of size $p^r$, we know that the translates of $T$ cover the whole of $G$, and therefore there are at least $n/p^r$ members in the orbit of $T$.
• Combining these two facts $S$ must have index exactly $m$ -- hence it is a subgroup.

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

Proof by conjugation action

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