Deducing basic facts about Sylow subgroups and Hall subgroups

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A sequel to this article, describing more advanced approaches to deducing facts about Sylow subgroups and Hall subgroups, is available at deducing advanced facts about Sylow subgroups and Hall subgroups

If you're interested in proving Sylow's theorem and its analogues, refer proving Sylow's theorem and its analogues. If you're interested in applications of Sylow numbers to prove that certain groups are not simple, refer using Sylow numbers to prove the existence of proper nontrivial normal subgroups

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 $H \le K \le G$, 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 subgroups and quotient maps

Further information: Sylow satisfies image condition, Hall satisfies image condition

Under a quotient map (or equivalently, under a surjective homomorphism), the image of a $p$-Sylow subgroup is $p$-Sylow in the image. The image of a $\pi$-Hall subgroup is $\pi$-Hall in the image.

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.

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 $gQg^{-1} \le P$. 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 $G$.

Equivalence of definitions of Sylow subgroup of normal subgroup

Further information: Equivalence of definitions of Sylow subgroup of normal subgroup

It turns out that a subgroup $P$ of $G$ is the intersection of a Sylow subgroup and a normal subgroup if and only if it is a Sylow subgroup of a normal subgroup. While one direction of proof involves the fact that Hall satisfies permuting transfer condition, i.e., an application of the product formula, the other direction involves also the use of the fact that given any $p$-subgroup, there exists a $p$-Sylow subgroup containing it. This is a combination of existence and domination: existence first guarantees the existence of some $p$-Sylow subgroup, while domination allows for conjugating this $p$-Sylow subgroup to a $p$-Sylow subgroup containing the specified $p$-subgroup.

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.

Resemblance properties for Sylow subgroups, with consequences

In this section, we use only two facts about Sylow subgroups:

• Sylow implies order-conjugate: A Sylow subgroup is conjugate to any other subgroup of the same order. This is the conjugacy part of Sylow's theorem.
• Sylow satisfies intermediate subgroup condition: A Sylow subgroup of the whole group is also Sylow in every intermediate subgroup.

A quick look at some equivalence relations between subgroups

Given two subgroups $H$ and $K$ of a finite group $G$, we can consider the following ways $H$ and $K$ may look similar:

1. Same order: $H$ and $K$ have the same order.
2. Isomorphic: $H$ and $K$ are isomorphic groups.
3. Automorphic:$H$ and $K$ are automorphic subgroups of $G$: there is an automorphism of $G$ that sends $H$ to $K$.
4. Conjugate: $H$ and $K$ are conjugate subgroups of $G$: there is an inner automorphism of $G$ that sends $H$ to $K$.
5. Equal: $H = K$.

Note that these are in increasing order of strength. Equal implies conjugate implies automorphic implies isomorphic implies same order.

Here's some terminology we'll use for a subgroup $H$ of $G$.

• Order-unique subgroup: (1) = (5). It is the only subgroup of its order in $G$.
• Order-conjugate subgroup: (1) = (4). Any subgroup of $G$ of the same order as $H$ is conjugate to it.
• Order-automorphic subgroup: (1) = (3).
• Order-isomorphic subgroup: (1) = (2).
• Isomorph-free subgroup: (2) = (5).
• Isomorph-conjugate subgroup: (2) = (4).
• Isomorph-automorphic subgroup: (2) = (3).
• Characteristic subgroup: (3) = (5).
• Automorph-conjugate subgroup: (3) = (4).
• Normal subgroup: (4) = (5).

For Sylow subgroups

For Sylow subgroups, we have (1) = (4). In other words, Sylow subgroups are order-conjugate subgroups. In particular, they satisfy a property $(m) = (n)$ for $1 \le m,n \le 4$. Thus, Sylow subgroups are order-automorphic, order-isomorphic, isomorph-conjugate, isomorph-automorphic, and automorph-conjugate.

For normal Sylow subgroups

For normal Sylow subgroups, we have (1) = (5). Thus, these subgroups satisfy all the properties listed above, and are also characteristic, isomorph-free, and order-unique. (This can actually be proved without using Sylow's theorem, and it generalizes to normal Hall subgroups, even though the analogue of Sylow's theorem fails there).

Intermediately each

We now use the fact that a Sylow subgroup of the whole group is also Sylow in every intermediate subgroup. Thus, we have the following:

• Sylow implies intermediately order-conjugate: A Sylow subgroup is an intermediately order-conjugate subgroup: it is order-conjugate in every intermediate subgroup.
• Sylow implies intermediately isomorph-conjugate: A Sylow subgroup is an intermediately isomorph-conjugate subgroup: it is isomorph-conjugate in every intermediate subgroup.
• Sylow implies intermediately automorph-conjugate: A Sylow subgroup is an intermediately automorph-conjugate subgroup: it is automorph-conjugate in every intermediate subgroup.

Procharacteristicity, pronormality, and a horde of related properties

The fact that any Sylow subgroup is intermediately isomorph-conjugate is the beginning of many exciting results about Sylow subgroups. We begin with a few definitions:

• Procharacteristic subgroup and pronormal subgroup: $H \le G$ is a procharacteristic subgroup if, for any automorphism $\sigma$ of $G$, $H$ and its image under $\sigma$ are conjugate in the subgroup they generate. $H$ is a pronormal subgroup of $G$ if, for any $g \in G$, $H$ and its conjugate by $g$ are conjugate in the subgroup they generate.
• Weakly procharacteristic subgroup and weakly pronormal subgroup: $H \le G$ is a weakly procharacteristic subgroup if, for any automorphism $\sigma$ of $G$, $H$ and its image under $\sigma$ are conjugate in the closure of $H$ in $G$ under the action of $\langle \sigma \rangle$. $H$ is weakly pronormal in $G$ if the above holds for all inner automorphisms $\sigma$.
• Paracharacteristic subgroup and paranormal subgroup: $H \le G$ is paracharacteristic if it is a contranormal subgroup inside the subgroup generated by $H$ and its image under any automorphism $\sigma$ of $G$. $H$ is paranormal if it is contranormal inside the subgroup generated by $H$ and its image under any inner automorphism. (A contranormal subgroup is a subgroup whose normal closure is the whole group).
• Polycharacteristic subgroup and polynormal subgroup: $H \le G$ is polycharacteristic if it is contranormal inside the closure in $G$ of $H$ under the action of $\langle \sigma \rangle$ for any automorphism $\sigma$. $H$ is polynormal in $G$ if this holds for any inner automorphism $\sigma$.

These pairs of properties are related as follows:

• Procharacteristic of normal implies pronormal, Left residual of pronormal by normal is procharacteristic
• Weakly procharacteristic of normal implies weakly pronormal, Left residual of weakly pronormal by normal is weakly procharacteristic
• Paracharacteristic of normal implies paranormal, Left residual of paranormal by normal is paracharacteristic
• Polycharacteristic of normal implies polynormal, Left residual of polynormal by normal is polycharacteristic

There's also a multitude of relations between these properties:

• Pronormal implies weakly pronormal, Procharacteristic implies weakly procharacteristic
• Pronormal implies paranormal, Procharacteristic implies paracharacteristic
• Weakly pronormal implies polynormal, Weakly procharacteristic implies polycharacteristic
• Paranormal implies polynormal, Paracharacteristic implies polycharacteristic

Here's how Sylow subgroups are related to all this. It turns out that:

• Intermediately isomorph-conjugate implies procharacteristic
• Intermediately automorph-conjugate implies weakly procharacteristic

In particular, since Sylow subgroups are intermediately isomorph-conjugate, they are procharacteristic. Thus, we have: Sylow implies procharacteristic, Sylow implies pronormal, Sylow of normal implies pronormal.

Subnormal-to-normal

Further information: Subnormal-to-normal and normal-to-characteristic

It turns out that each of the properties: procharacteristic, weakly procharacteristic, paracharacteristic, and polycharacteristic, are such that any normal subgroup with the property is also a characteristic subgroup. Thus, these are all stronger than the property of being a normal-to-characteristic subgroup. In particular, any normal Sylow subgroup is characteristic (we already knew this).

It also turns out that the properties: pronormal, weakly pronormal, paranormal, and polynormal, are all stronger than the property of being an intermediately subnormal-to-normal subgroup: any subgroup with any of these properties that is subnormal in some intermediate subgroup is normal in that intermediate subgroup. In particular, any Sylow subgroup of a normal subgroup, being pronormal, is intermediately subnormal-to-normal: if it is subnormal in some intermediate subgroup, it is also normal in that intermediate subgroup.

Frattini's argument

Further information: Frattini's argument

It might be worth making a quick remark about Frattini's argument. In its most general form, it says that if $H \le M \le G$ such that $H$ is an automorph-conjugate subgroup of $M$ and $M$ is normal in $G$, then $MN_G(H) = G$. Since, as discussed earler, Sylow subgroups are automorph-conjugate, we can replace automorph-conjugate above by Sylow.

About Hall subgroups

Failure of existence, conjugacy and domination

Existence fails: Hall subgroups need not exist. In fact, Hall subgroups of all possible orders exist if and only if the group is solvable, according to Hall's theorem.

Conjugacy fails rather spectacularly, as we shall see in the next section.

Domination fails, even if we assume that all the Hall subgroups of that order are conjugate: Order-conjugate and Hall not implies order-dominating.

Domination in the reverse sense also fails, even if we assume that all the Hall subgroups of that order are conjugate: Order-conjugate and Hall not implies order-dominated.

Resemblance notions usually fail for Hall subgroups

Practically all the relation implication properties that are true for Sylow subgroups break down for Hall subgroups. Here are some of the breakdowns:

• Hall not implies order-isomorphic: Two Hall subgroups of a finite group of the same order (and thus, correpsonding to the same prime set) need not be isomorphic.
• Hall not implies isomorph-automorphic: Two isomorphic Hall subgroups of a finite group need not be automorphic subgroups.
• Hall not implies automorph-conjugate: Two Hall subgroups of a finite group may be related via an outer automorphism but not be conjugate subgroups.

Hall subgroups are paracharacteristic

Can we use the good behavior of Sylow subgroups to deduce good behavior for Hall subgroups? We can use the fact that Hall implies join of Sylow subgroups. The problem is that most of the properties mentioned here are not closed under arbitrary joins -- they are closed only under more restrictive kinds of joins such as normalizing joins.

For instance, it turns out that joins of procharacteristic subgroups need not be procharacteristic, and Hall subgroups need not be procharacteristic.

However, joins of paracharacteristic subgroups are paracharacteristic. Since Sylow subgroups are procharacteristic, procharacteristic subgroups are paracharacteristic, and Hall subgroups are joins of Sylow subgroups, Hall subgroups are always paracharacteristic. In particular, this yields that Hall subgroups of normal subgroups are paranormal.

In particular, any normal Hall subgroup is characteristic, and any subnormal subgroup that is also a Hall subgroup of a normal subgroup must be normal in the whole group.

Nilpotent Hall subgroups and normalizing joins

A nilpotent Hall subgroup is a Hall subgroup that is also a nilpotent group. Since groups of prime power order are nilpotent, all Sylow subgroups are nilpotent Hall subgroups. It turns out that nilpotent Hall subgroups behave very similarly to Sylow subgroups.

To understand this better, define a subgroup property $p$ to be normalizing join-closed if, whenever $H, K \le G$ both satisfy $p$ and $K \le N_G(H)$, $HK$ also satisfies $p$. Many of the properties discussed here are normalizing join-closed; for instance:

• Intermediate isomorph-conjugacy is normalizing join-closed
• Intermediate automorph-conjugacy is normalizing join-closed
• Pronormality is normalizing join-closed
• Procharacteristicity is normalizing join-closed
• Weak pronormality is normalizing join-closed
• Weak procharacteristicity is normalizing join-closed

Since a finite nilpotent group has all Sylow subgroups normal, it can be expressed in terms of normalizing joins, one by one, of Sylow subgroups. Thus, nilpotent Hall subgroups satisfy all these properties:

• Nilpotent Hall implies intermediately isomorph-conjugate
• Nilpotent Hall implies intermediately automorph-conjugate
• Nilpotent Hall implies procharacteristic
• Nilpotent Hall of normal implies pronormal

In fact, something slightly stronger is true for nilpotent Hall subgroups, namely:

Nilpotent Hall subgroups of same order are conjugate

Note that this is stronger than saying that nilpotent Hall subgroups are isomorph-conjugate (which we already claimed to be true) but is weaker than saying that nilpotent Hall subgroups are order-conjugate (which is false).

Analogue of Sylow's theorem for Hall subgroups in finite solvable groups

The quick statement

Further information: Hall subgroups exist in finite solvable, Hall implies order-conjugate in finite solvable, Hall implies order-dominating in finite solvable, ECD condition for pi-subgroups in solvable groups

In a finite solvable group, Hall subgroups of all possible orders exist, any two $\pi$-Hall subgroups are conjugate, and any $\pi$-subgroup is contained in a $\pi$-Hall subgroup. Further, since subgroups and quotients of solvable groups are solvable, the conditions of existence, conjugacy, and domination also hold in all intermediate subgroups and quotients. There's also a part on the number of Hall subgroups, but this is rather complicated.

Thus, everything that we said in the preceding section about Sylow subgroups in general holds for Hall subgroups in solvable groups. In particular, for instance, Hall of solvable normal implies pronormal: any Hall subgroup of a solvable normal subgroup is pronormal.

The proofs of existence, conjugacy, and domination rely crucially on the corresponding parts for Sylow's theorem, the fact that subgroups and quotients of solvable groups are solvable, the result called Frattini's argument (which essentially follows from the conjugacy part of Sylow's theorem), and the following crucial fact about finite solvable groups: minimal normal implies elementary abelian in finite solvable.

A converse

Further information: Hall's theorem

If $\pi$-Hall subgroups exist for all prime sets $\pi$, then the group is solvable.