Sylow Theorems

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

$p$-Groups

In this post, $p$ always denotes a prime number.

Definition 1 A finite group $G$ is called a $p$-group if the order of $G$ is a power of $p$.

Then it is obvious that every subgroup and quotient group of a $p$-group is again a $p$-group. Moreover, the following holds.

Lemma 2 Let a $p$-group $G$ act on a finite set $E$, and consider the set of fixed points of this action

\[E^G=\{x\in E\mid g\cdot x=x\text{ for all $g\in G$}\}.\]

Then

\[\lvert E^G\rvert\equiv\lvert E\rvert\pmod{p}\]

holds.

Proof

That is, we must show that the size of $E\setminus E^G$ is a multiple of $p$. Now $E\setminus E^G$ is a union of (disjoint) $G$-orbits each of size greater than $1$, and by §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem) each such orbit has size a power of $p$, whence the claim.

In particular, if we consider the case $E=G$ with $G$ acting by inner automorphisms, then $E^G$ is precisely the center of $G$; thus by Lemma 2 the center $C(G)$ of a $p$-group $G$ is nontrivial.

Theorem 3 For a $p$-group $G$ of order $p^r$, there exists a series of subgroups

\[G=G_1\supset G_2\supset\cdots G_{n+1}=\{e\}\]

such that $[G, G_k]\subseteq G_{k+1}$ holds for every $k$, and each $G_k/G_{k+1}$ is a cyclic group of order $p$.

Proof

We prove this by induction on the order of $G$. If $G=\{e\}$ there is nothing to prove. Assume the statement holds for all $p$-groups of order less than $\lvert G\rvert=p^r$, and let us prove it for $\lvert G\rvert=p^r$. From the previous argument we have $C(G)\neq\{e\}$, so we can choose an element $x\in C(G)$ of order $p^s$ ($1\leq s\leq r$).

Now let $H$ be the subgroup of $C(G)$ generated by the element $x^{p^{s-1}}$. Then $G'=G/H$ is a $p$-group of order $p^{r-1}$, so by the inductive hypothesis there exists a series of subgroups satisfying the given condition, and taking the inverse image under the canonical projection $p: G \rightarrow G'$ yields the desired series.

Therefore, by the equivalence of the first and second conditions in §Series of a Group, ⁋Proposition 7, every $p$-group is nilpotent.

On the other hand, from §Series of a Group, ⁋Proposition 8 we obtain the following.

Proposition 4 Let $G$ be a $p$-group and let $H\subsetneq G$ be a subgroup.

The normalizer $N_G(H)$ of $H$ in $G$ satisfies $N_G(H)\subsetneq G$.
There exists a normal subgroup $N$ of $G$ of index $p$ containing $H$.

Hence every subgroup of index $p$ in a $p$-group $G$ is normal.

Sylow Theorems

We now examine $p$-subgroups of a general group. Among them, the following are of particular interest.

Definition 5 A Sylow $p$-subgroup of a finite group $G$ is a subgroup $P$ of $G$ satisfying the following two conditions:

$P$ is a $p$-group.
$[G:P]$ is not divisible by $p$.

We write $\Syl_p(G)$ for the set of Sylow $p$-subgroups of $G$.

Thus, when the order of $G$ is given as $p^r m$ with $p\nmid m$, a Sylow $p$-subgroup $P$ is precisely a subgroup of $G$ of order $p^r$. For the remainder of this post we always assume that the order $n=p^rm$ of $G$ satisfies $p\nmid m$.

The Sylow theorems are theorems about Sylow $p$-subgroups of an arbitrary finite group, and they are useful in classifying finite groups. The first result concerns the existence of a Sylow $p$-subgroup; for this we need the following lemma.

Lemma 6 Let $n=p^rm$ with $p\nmid m$. Then

\[\binom{n}{p^r}\not\equiv 0\pmod{p}\]

holds.

Proof

Consider a group $G$ of order $p^r$ and a set $S$ of size $m$. Then the set $G\times S$ has size $n$; let $E$ be the set of subsets of $G\times S$ of size $p^r$, so that

\[\lvert E\rvert=\binom{n}{p^r}.\]

If $G$ acts on $G\times S$ by

\[g \cdot (x, s) = (g x, s) \quad (g, x \in G,\; s \in S),\]

then by applying this action to each element of every element of $E$ (that is, to each element of each subset of $G\times S$ of size $p^r$) we obtain an action of $G$ on $E$. The set of fixed points $E^G$ of this action consists exactly of subsets of the form

\[G \times \{s\},\qquad s\in S,\]

so $\lvert E^G\rvert=m$, and now by Lemma 2

\[\binom{n}{p^r} = \text{Card}(E) \equiv \text{Card}(E^G) = m \not\equiv 0 \pmod{p}\]

holds.

The first Sylow theorem is then the existence of a Sylow $p$-subgroup.

Theorem 7 $G$ has a Sylow $p$-subgroup.

Proof

Let $E$ be the set of subsets of $G$ having exactly $p^r$ elements. Then by Lemma 6

\[\lvert E\rvert = \binom{n}{p^r}\not\equiv 0\pmod{p}.\]

Now consider the left translation action of $G$ on itself

\[L_g:G \rightarrow G;\qquad x\mapsto gx\]

and view this as an action on $E$ in the same way as in the proof of Lemma 6. From the assumption that $\lvert E\rvert\not\equiv 0\pmod{p}$, there exists an orbit $O$ whose size is not a multiple of $p$. Let $X$ be an element of $O$, and let the stabilizer of $X$ be $\Stab(\{X\})=\Stab(X)$. Then $\Stab(X)$ is a subgroup of $G$, and (§Group Actions, ⁋Corollary 8) this is the subgroup we want.

To see this, we first use §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem) to obtain

\[\lvert O\rvert=\lvert G\cdot X\rvert=[G:\Stab(X)]=\frac{\lvert G\rvert}{\lvert\Stab(X)\rvert}\not\equiv 0\pmod{p},\]

so $p^r$ divides $\lvert \Stab(X)\rvert$.

On the other hand, $\Stab(X)$ is the set of elements $g\in G$ satisfying $gX = X$, and therefore for any element $x \in X$ we have

\[\Stab(X) \subseteq X x^{-1},\]

\[\lvert \Stab(X)\rvert\leq\lvert Xx^{-1}\rvert=\lvert X\rvert=p^r\]

must hold. From this we conclude that $\lvert\Stab(X)\rvert=p^r$.

From this we know that if the order of an arbitrary finite group $G$ is divisible by $p$, then $G$ has an element of order $p$.

For two subgroups $H_1, H_2$ of $G$, we say that $H_1$ and $H_2$ are conjugate if there exists $\rho\in\Inn(G)$ such that $\rho(H_1)=H_2$.

Theorem 8 The following hold.

The Sylow $p$-subgroups of $G$ are conjugate to each other, and their number is $1$ mod $p$.
Every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.

Proof

Let $P$ be a Sylow $p$-subgroup of $G$, and let $H$ be a $p$-subgroup of $G$. Considering the left translation action of $H$ on the set $E = G/P$, we have $\lvert E^H\rvert\neq 0$ by Lemma 6, so there exists $x\in G/P$ with $Hx=x$. Choose a representative $g\in G$ of the element $x$ of $G/P$. Then for any $h \in H$ we have $h(gP) = gP$, so $g^{-1} h g \in P$. Hence $H \subseteq gPg^{-1}$, which proves the second claim.

Now suppose $H$ is a Sylow $p$-subgroup. Then

\[\lvert H \rvert = \lvert P \rvert = \lvert gPg^{-1} \rvert,\]

so the above inclusion becomes $H = gPg^{-1}$, proving the first part of the first claim.

To prove the second part of the first claim, let $G$ act on $\Syl_p(G)$ by inner automorphisms. Then from the previous argument any $P \in \Syl_p(G)$ is a fixed point of this action, and we show that it is the unique fixed point.

Suppose for contradiction that there is another fixed point $Q \in \Syl_p(G)$. Then $Q$ is a Sylow $p$-subgroup of $G$ normalized by $P$; that is, $P\subseteq N_G(Q)$. Now $P$ and $Q$ are both Sylow $p$-subgroups of $N_G(Q)$, so by the previous argument there exists $n \in N_G(Q)$ such that

\[P = nQn^{-1} = Q\]

holds. Therefore by Lemma 6 we have $\lvert \Syl_p(G) \rvert = \lvert \Syl_p(G)^P \rvert \equiv 1 \pmod{p}$.

Corollary 9 Let $P\in\Syl_p(G)$ and consider its normalizer $N_G(P)$. For any subgroup $M$ of $G$ containing $N_G(P)$, the normalizer $N_G(M)$ of $M$ in $G$ equals $M$.

Proof

Choose $g\in G$ satisfying $M=gMg^{-1}$. Then $gPg^{-1}$ is a Sylow $p$-subgroup of $M$. Hence there exists $h \in M$ such that $gPg^{-1} = hPh^{-1}$. Now $h^{-1}g \in N$, and therefore $g \in hN \subset M$.

Corollary 10 Let $f: G_1 \to G_2$ be a group homomorphism between finite groups. For any Sylow $p$-subgroup $P_1$ of $G_1$, there exists a Sylow $p$-subgroup $P_2$ of $G_2$ containing $f(P_1)$.

Proof

Apply the second result of Theorem 8 to the subgroup $f(P_1)$ of $G_2$.

Corollary 11 1. Let $H$ be a subgroup of $G$. For any Sylow $p$-subgroup $P$ of $H$, there exists a Sylow $p$-subgroup $Q$ of $G$ such that $P = Q \cap H$.

Conversely, let $Q$ be a Sylow $p$-subgroup of $G$ and let $H$ be a normal subgroup of $G$. Then $Q \cap H$ is a Sylow $p$-subgroup of $H$.

Proof

The $p$-group $P$ is contained in a Sylow $p$-subgroup $Q$ of $G$. On the other hand $Q \cap H$ is a $p$-subgroup of $H$ containing $P$, so eventually $P = Q \cap H$.
Let $P'$ be a Sylow $p$-subgroup of $H$. Then there exists $g \in G$ such that $gP'g^{-1} \subset Q$. Since $H$ is normal, $P = gP'g^{-1}$ is again contained in $H$, and therefore $P$ is contained in $Q\cap H$. Now $Q \cap H$ is a $p$-subgroup of $H$, and $P$ is a Sylow $p$-subgroup, so $P = Q \cap H$.

Corollary 12 Let $N$ be a normal subgroup of $G$. Then the image in $G/N$ of any Sylow $p$-subgroup of $G$ is a Sylow $p$-subgroup of $G/N$, and moreover every Sylow $p$-subgroup of $G/N$ arises in this way.

Proof

Fix $P\in \Syl_p(G)$, let $G' = G/N$, and let $P'$ be the image of $P$ in $G'$.

Considering the left translation action of $G$ on $G'/P'$, this is a transitive action, so the orbit of $G$ is $G'/P'$ itself. Now by §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem)

\[\lvert G'/P'\rvert=[G:\Stab(G'/P')]\]

holds. But by definition $\Stab(G'/P')$ contains $P$, so $[G:\Stab(G'/P')]$ is not divisible by $p$, and hence $[G':P']$ is also not divisible by $p$. On the other hand $P'$ is a $p$-group, so by definition $P'$ is a Sylow $p$-subgroup of $G'$.

For the converse, let $Q'$ be another Sylow $p$-subgroup of $G'$. Then $Q' = g'P'g'^{-1}$ for some $g' \in G'$, and taking a representative $g \in G$ of $g'$, the image of $gPg^{-1}$ is $Q'$.

Applications of the Sylow Theorems

As mentioned above, the Sylow theorems are useful in classifying finite groups. For this purpose let us examine Theorem 8 a little more closely. Writing $n_p$ for the size of $\Syl_p(G)$, the second part of the first result of Theorem 8 gives $n_p\equiv 1\pmod{p}$. On the other hand, the first part of the first result shows that $G$ acts transitively on $\Syl_p(G)$, so by §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem)

\[n_p=\lvert \Syl_p(G)\rvert=[G:\Stab(P)],\qquad P\in\Syl_p(G)\]

holds; in particular $n_p$ must divide $\lvert G\rvert$, and as we saw above $n_p$ does not divide $p^r$, so $n_p$ must divide $m$.

Example 13 Let us classify finite groups $G$ of order $15$.

\[\lvert G\rvert = 15 = 3\times 5.\]

First consider the Sylow 3-subgroups. Then by Theorem 8 the number $n_3$ of Sylow 3-subgroups satisfies the two conditions

$n_3\equiv 1\pmod{3}$,
$n_3$ divides $5$.

The only $n_3$ satisfying these two conditions is $1$, and by Theorem 8 this means that the (unique) Sylow $3$-subgroup $P_3$ of $G$ is normal.

Similarly consider the Sylow 5-subgroups. By the Sylow theorems, the number $n_5$ of Sylow 5-subgroups satisfies:

$n_5\equiv 1\pmod{5}$,
$n_5$ divides $3$.

Likewise the only $n_5$ satisfying these two conditions is $1$, so the Sylow 5-subgroup also exists uniquely and is normal. Denote it by $P_5$.

Now $P_3\cap P_5$ is a subgroup of both $P_3$ and $P_5$, so its order must be $1$, and hence $P_3\cap P_5=\{e\}$. Considering the subgroup $P_3P_5$ of $G$, from §Isomorphism Theorems, ⁋Theorem 5 (The Second Isomorphism Theorem)

\[\frac{P_3P_5}{P_3}\cong P_5/\{e\}\implies \lvert P_3P_5\rvert=\lvert P_3\rvert\lvert P_5\rvert=15=\lvert G\rvert\]

so eventually $G\cong \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$.

Share on

Twitter Facebook LinkedIn

K

Sylow Theorems

\(p\)-Groups

Sylow Theorems

Applications of the Sylow Theorems

Share on

댓글남기기