Deducing basic facts about Sylow subgroups and Hall subgroups

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This is a survey article related to:Hall subgroup
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This survey article explores how we can deduce various facts about Sylow subgroups. The theme here is as follows: we state some important theorems, including the components of Sylow's theorem, as well as some basic facts about Sylow subgroups, and then use these as black box theorems to deduce a number of powerful corollaries about the way Sylow subgroups and Hall subgroups look and behave.

Basic definitions

Hall subgroup

Further information: Hall subgroup

A subgroup $H$ of a finite group $G$ is termed a Hall subgroup if the order and index of $H$ are relatively prime.

In terms of prime factorization, this is expressible as follows. Suppose the order of $G$ is:

$n=p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{r}^{k_{r}}$.

Then, a subgroup $H$ of $G$ is a subgroup whose order is of the form:

$\prod _{i\in S}p_{i}^{k_{i}}$,

where $S$ is some subset of $\{1,2,3,\dots ,r\}$. Note that the empty subset corresponds to the trivial group and the whole set corresponds to the whole group.

If $\pi$ is a set of primes such that all prime divisors of the order of $H$ are in $\pi$ but no prime divisors of the order of $H$ are in $\pi$, then $H$ is termed a $\pi$-Hall subgroup. Note that $\pi$ must include all the primes $p_{i},i\in S$, but it can also include other primes that do not divide the order of $G$. For instance, a subgroup of order $12$ in a group of order $60$ is a $\{2,3\}$-Hall subgroup, but it is also a $\{2,3,7\}$-Hall subgroup.

Sylow subgroup

Further information: Sylow subgroup

Let $G$ be a finite group and $p$ be a prime. A $p$-Sylow subgroup of $G$ is a Hall subgroup whose order is a power of $p$. Equivalently, it is a subgroup whose order is the largest power of $p$ dividing the order of $G$.

Note that if $p$ is relatively prime to the order of $G$, then the trivial subgroup is the unique $p$-Sylow subgroup of $G$.

A Sylow subgroup is a subgroup that is $p$-Sylow for some prime $p$.

Pre-Sylow's theorem facts about Sylow subgroups and Hall subgroups

These are facts whose proof requires basic ideas about order and index and a use of the facts regarding subgroups and cosets.

Hall satisfies transitivity

Further information: Hall satisfies transitivity

A Hall subgroup of a Hall subgroup is a Hall subgroup. This follows from the fact that index is multiplicative: if $H\leq K\leq G$, we have:

$[G:H]=[G:K][K:H]$.

Sylow and Hall subgroups in intermediate subgroups

Further information: Sylow satisfies intermediate subgroup condition, Hall satisfies intermediate subgroup condition

These are basic facts, again following from multiplicativity of index and/or Lagrange's theorem:

• In a given finite group, any two $p$-Sylow subgroups have the same order, and any two $\pi$-Hall subgroups have the same order.
• Hall satisfies intermediate subgroup condition: Suppose $H$ is a Hall subgroup of $G$, and $K$ is a subgroup of $G$ containing $H$. Then, $H$ is also a Hall subgroup of $K$. More specifically, if $H$ is $\pi$-Hall in $G$, $H$ is $\pi$-Hall in $K$.
• Sylow satisfies intermediate subgroup condition: If $P$ is a $p$-Sylow subgroup of $G$, $P$ is also $p$-Sylow in every intermediate subgroup.
• Suppose $H$ is a $\pi$-Hall subgroup of $G$ and $K$ is a subgroup whose order is divisible by the order of $G$. Then, if $K$ has a $\pi$-Hall subgroup, the order of that $\pi$-Hall subgroup is the same as the order of $H$.

Sylow and Hall satisfy permuting transfer condition

Further information: Hall satisfies permuting transfer condition

Suppose $H$ is a Hall subgroup of $G$, and $K$ is a subgroup of $G$ such that $HK=KH$, i.e., $H$ and $K$ are permuting subgroups. This happens, for instance, if either subgroup normalizes the other. Then, $H\cap K$ is a Hall subgroup of $K$. Specifically, if $H$ is $\pi$-Hall in $G$, $H\cap K$ is also $\pi$-Hall in $K$.

The proof is an application of the product formula, which is a coset space version of the second isomorphism theorem.

(Note that later results will show that if $HK=KH$, then in fact $H$ is normal in $HK$; however, we do not need this for the proof).

Some consequences of this:

• Sylow satisfies permuting transfer condition: The analogous result for Sylow subgroups follows.
• Intersection of Hall subgroup and normal subgroup implies Hall subgroup of normal subgroup: This follows pretty easily.
• Equivalence of definitions of Sylow subgroup of normal subgroup: One direction follows the same way as the previous step. The other direction requires using the fact that Sylow subgroups exist.

The components of Sylow's theorem

We're now in a position to look at the four major components of Sylow's theorem.

Sylow subgroups exist

Further information: Sylow subgroups exist

If $G$ is a finite group and $p$ is a prime number, then $G$ has a $p$-Sylow subgroup. Note that when $p$ does not divide the order of $G$, the trivial subgroup is the unique $p$-Sylow subgroup, so the statement provides information only when $p$ does divide the order of $G$.

Sylow implies order-dominating

Further information: Sylow implies order-dominating

A $p$-Sylow subgroup of a group $G$ is a subgroup whose order is the largest power of $p$ dividing the order of $G$. In particular, this means that if $P$ is $p$-Sylow in $G$ and $Q$ is a $p$-subgroup of $G$, then the order of $Q$ divides the order of $P$.

The domination part of Sylow's theorem states that, in fact, some conjugate of $Q$ is contained in $P$. In other words, there exists $g\in G$ such that $gQg^{-1}\leq P$. Note that this is equivalent to saying that any subgroup of $G$ whose order divides the order of $P$ is contained in some conjugate of $P$.

Sylow implies order-conjugate

Further information: Sylow implies order-conjugate

As remarked earlier, any two $p$-Sylow subgroups have the same order. Conversely, any subgroup of $G$ of the same order as a $p$-Sylow subgroup is also a $p$-Sylow subgroup.

The previous domination fact about $p$-Sylow subgroups yields the following: any two $p$-Sylow subgroups of $G$ are conjugate. In particular, a $p$-Sylow subgroup is conjugate to any subgroup of the same order.

Conditions on Sylow numbers

Further information: Congruence condition on Sylow numbers, Divisibility condition on Sylow numbers

The $p$-Sylow number of a group, denoted $n_{p}$, is defined as the number of $p$-Sylow subgroups of $G$. The congruence condition on Sylow numbers states that $n_{p}$ is $1$ modulo $p$, while the divisibility condition states that $n_{p}$ divides the Sylow index: the index of any $p$-Sylow subgroup.

The congruence conditions on Sylow numbers will not be used for the bulk of this article.

Combining existence with conjugacy/domination

We now give a few applications on how to combine existence of Sylow subgroups with conjugacy/domination, and the facts mentioned about Sylow and Hall subgroups in intermediate subgroups. It turns out that the fact that Sylow subgroups of a group remain Sylow subgroups in intermediate subgroups is of profound importance.

Hall subgroups are joins of Sylow subgroups

Further information: Hall implies join of Sylow subgroups

Suppose $H$ is a $\pi$-Hall subgroup of $G$. Then, $H$ is a join of $p$-Sylow subgroups of $G$, $p\in \pi$.

The proof of this involves the existence of $p$-Sylow subgroups in $H$, arguing that $H$ is the join of these, and then arguing that since Hall satisfies transitivity, these $p$-Sylow subgroups in $H$ are also $p$-Sylow in $G$.

Sylow implies order-dominated

Further information: Sylow implies order-dominated

If $P$ is a $p$-Sylow subgroup of a finite group $G$ and $H$ is a subgroup of $G$ such that the order of $P$ divides the order of $H$, then some conjugate of $P$ is contained in $H$.

The proof of this relies on the fact that $H$ itself has a $p$-Sylow subgroup, followed by the fact that this is also a $p$-Sylow subgroup of $G$, followed by the fact that any two $p$-Sylow subgroups of $G$ are conjugate.