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.

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

It is then obvious that subgroups and quotient groups of a $p$-group are again $p$-groups. 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 $p$. But $E\setminus E^G$ is the union of (disjoint) $G$-orbits each of size greater than $1$, and the size of each such orbit is a power of $p$ by [Algebraic Structures] §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem), so this holds.

In particular, considering the case where $E=G$ and $G$ acts by inner automorphisms, $E^G$ is exactly the center of $G$, so by Lemma 2 we know that the center $Z(G)$ of a $p$-group $G$ is not the trivial group.

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

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

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

Proof

We prove this by induction on the order of $G$. The case $G=\{e\}$ is trivial. Now assume that the given statement holds for all $p$-groups of order less than $\lvert G\rvert=p^r$, and let us prove the case $\lvert G\rvert=p^r$. From the previous argument, $Z(G)\neq\{e\}$, so there exists some $x\in Z(G)$ whose order is $p^s$ ($1\leq s\leq r$).

Now consider the subgroup $H$ of $Z(G)$ 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 of §Series of a Group, ⁋Proposition 7, we know that every $p$-group is nilpotent.

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

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

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 $H$.

Therefore, 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 denote the set of Sylow $p$-subgroups of $G$ by $\Syl_p(G)$.

That is, when the order of $G$ is given as $p^r m$ with $p\nmid m$, a Sylow $p$-group $P$ 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 help in classifying finite groups. The first result concerns the existence of Sylow $p$-subgroups, and 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 $G\times S$ is a set of size $n$, and defining $E$ to be the set of subsets of $G\times S$ of size $p^r$, we have

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

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

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

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

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

holds.

Then the first result of the Sylow theorems is the existence of Sylow $p$-subgroups.

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

Proof

Let $E$ be the set of subsets of $G$ with $p^r$ 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 action as an action on $E$ in the same way as in the proof of Lemma 6. Then from the assumption $\lvert E\rvert\not\equiv 0\pmod{p}$, there exists an orbit $O$ whose size is not a multiple of $p$. Now 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 by ([Algebraic Structures] §Group Actions, ⁋Corollary 8) this is the subgroup we want.

To show this, we first obtain from [Algebraic Structures] §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem)

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

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

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

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

and therefore

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

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 $E = G/P$, by Lemma 2 we have $\lvert E^H\rvert\neq 0$, so there exists $x\in G/P$ with $Hx=x$. Now 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 $g^{-1} h g \in P$. Therefore $H \subseteq gPg^{-1}$, and this 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 this is the only 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 2 we know that $\lvert \Syl_p(G) \rvert \equiv \lvert \Syl_p(G)^P \rvert = 1 \pmod{p}$.

Corollary 9 Let $P\in\Syl_p(G)$ and consider its normalizer $N_G(P)$. For a 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 $M$. Therefore there exists $h \in M$ such that $gPg^{-1} = hPh^{-1}$. Now $h^{-1}g \in N_G(P)$ and therefore $g \in hN_G(P) \subseteq M$.

Corollary 10 Fix a group homomorphism $f: G_1 \rightarrow G_2$ between finite groups. For a 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

Let $H$ be a subgroup of $G$. For a 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 $gP'g^{-1} \subseteq Q$. Since $H$ is a normal subgroup, $P = gP'g^{-1}$ 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 of a Sylow $p$-subgroup of $G$ in $G/N$ is a Sylow $p$-subgroup of $G/N$, and moreover every Sylow $p$-subgroup of $G/N$ is obtained 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 [Algebraic Structures] §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem)

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

But by definition $\Stab(G'/P')$ contains $P$, so $[G:\Stab(G'/P')]$ is not divisible by $p$, and therefore $[G':P']$ 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, if we consider another Sylow $p$-subgroup $Q'$ of $G'$, then for some $g' \in G'$ we have $Q' = g'P'g'^{-1}$, and taking a representative $g \in G$ of $g'$, the image of $gPg^{-1}$ becomes $Q'$.

Applications of the Sylow Theorems

As mentioned earlier, the Sylow theorems are useful for the classification of finite groups. To this end, let us examine Theorem 8 more closely. Let $n_p$ be the size of $\Syl_p(G)$. By the second part of the first result of Theorem 8, we have $n_p\equiv 1\pmod{p}$. On the other hand, the first part of the first result of Theorem 8 shows that $G$ acts transitively on $\Syl_p(G)$, so by [Algebraic Structures] §Group Actions, ⁋Theorem 14 (Orbit-stabilizer theorem)

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

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

Example 13 Let us classify finite groups 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 following two conditions:

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

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

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. Let us 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 therefore $P_3\cap P_5=\{e\}$. Now considering the subgroup $P_3P_5$ of $G$, by [Algebraic Structures] §Group Homomorphisms, ⁋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 $G\cong \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$.

As another powerful application of the Sylow theorems, we show that the alternating group $A_5$ is a simple group. This was already proved in §Symmetric Groups, ⁋Example 13 by directly investigating the structure of conjugacy classes. There, the fact that $A_5$ is not such a large group was used to classify the types of elements one by one; here we derive the same conclusion by counting the Sylow subgroups that a normal subgroup must contain.

Proposition 14 The alternating group $A_5$ is a simple group.

Proof

The order of the alternating group is

\[\lvert A_5\rvert=\frac{5!}{2}=60=2^2\cdot 3\cdot 5\]

Also, as we saw in §Symmetric Groups, ⁋Example 13, the elements of $A_5$ consist of $1$ identity element, $20$ elements of order $3$ (3-cycles), $15$ elements of order $2$ (double transpositions), and $24$ elements of order $5$ (5-cycles).

Now suppose $N\triangleleft A_5$ satisfies $N\neq\{e\}$, and let us show that $N=A_5$. The order $\lvert N\rvert$ divides $60$, and among the divisors of $60$, those not divisible by the primes $3$ or $5$ are only $1,2,4$. Therefore either $3$ or $5$ divides $\lvert N\rvert$, or else $\lvert N\rvert\in\{2,4\}$.

First suppose $5\mid\lvert N\rvert$. Then $N$ has an element of order $5$, and the subgroup $P$ generated by this element is a Sylow 5-subgroup of $A_5$. Since $N$ is normal, for any $g\in A_5$ we have $gPg^{-1}\subseteq gNg^{-1}=N$, and by Theorem 8 every Sylow 5-subgroup of $A_5$ is conjugate to $P$, so eventually $N$ contains all of them. On the other hand, there are $24$ 5-cycles in $A_5$, each Sylow 5-subgroup contains exactly $4$ 5-cycles, and distinct Sylow 5-subgroups are subgroups of order $5$ so they share only the identity element. Therefore the number of Sylow 5-subgroups is $n_5=24/4=6$, which is consistent with the Sylow theorem results $n_5\equiv 1\pmod{5}$ and $n_5\mid 12$. Thus $N$ contains $24$ elements of order $5$, so $\lvert N\rvert\geq 25$, and since $5\mid\lvert N\rvert$ and $\lvert N\rvert\mid 60$, we must have $\lvert N\rvert\in\{30,60\}$.

Similarly, in the case $3\mid\lvert N\rvert$, by the same argument we know that $N$ contains all Sylow 3-subgroups of $A_5$. There are $20$ 3-cycles, each Sylow 3-subgroup has $2$ 3-cycles, and distinct ones share only the identity element, so $n_3=20/2=10$, and $N$ contains $20$ elements of order $3$, so $\lvert N\rvert\geq 21$. Likewise $\lvert N\rvert\in\{30,60\}$.

Now let us exclude the case $\lvert N\rvert\in\{2,4\}$. If $\lvert N\rvert=4$, then $N$ is a Sylow 2-subgroup of $A_5$, and since it is normal, it is the unique Sylow 2-subgroup of $A_5$. Every element of order $2$ belongs to some Sylow 2-subgroup through the subgroup it generates, so if such a subgroup is unique, all $15$ elements of order $2$ must lie inside it. However, $A_5$ has no elements of order $4$ (the possible orders are only $1,2,3,5$), so each Sylow 2-subgroup is a Klein four-group $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, and therefore has exactly $3$ elements of order $2$. This is a contradiction. On the other hand, if $\lvert N\rvert=2$, then $N=\langle x\rangle$ and $x$ is a double transposition; for $N$ to be normal, we must have $gxg^{-1}=x$ for any $g\in A_5$, that is, $x$ must be central. But conjugating an arbitrary double transposition $(ab)(cd)$ by the 3-cycle $(abc)$ gives $(abc)(ab)(cd)(abc)^{-1}=(ad)(bc)$, which is different from the original element, so a double transposition cannot be central. This is also a contradiction. Therefore $\lvert N\rvert\notin\{2,4\}$, and by the previous argument $\lvert N\rvert\in\{30,60\}$.

Finally, let us exclude the case $\lvert N\rvert=30$. Since $3\mid 30$ and $5\mid 30$, by the first two arguments $N$ must contain all $20$ elements of order $3$ and all $24$ elements of order $5$. These have different orders so they do not overlap, and including the identity element, $N$ has at least $20+24+1=45$ elements. This contradicts $\lvert N\rvert=30$.

The only remaining possibility is $\lvert N\rvert=60$, that is, $N=A_5$. Therefore $A_5$ is a simple group.

References

[DF] D. S. Dummit and R. M. Foote, Abstract algebra, 3rd ed., Wiley, 2004.

Share on

Twitter Facebook LinkedIn

K

Sylow Theorems

\(p\)-Groups

Sylow Theorems

Applications of the Sylow Theorems

Share on

댓글남기기

찾아보기