# Sylow Theorems

The **Sylow theorems** are important tools for analysis of special subgroups of a finite group $G,$ known as **Sylow subgroups**. They are especially useful in the classification of finite simple groups.

The first Sylow theorem guarantees the existence of a Sylow subgroup of $G$ for any prime $p$ dividing the order of $G.$ A Sylow subgroup is a subgroup whose order is a power of $p$ and whose index is relatively prime to $p.$ The second Sylow theorem states that all the Sylow subgroups of a given order are conjugate, and the third Sylow theorem gives information about the number of Sylow subgroups.

The proofs of the theorems use nontrivial facts about group actions, in particular the action of $G$ on the coset space $G/H,$ where $H$ has prime power order.

## The Theorems

Let $G$ be a finite group. Let $p$ be a prime dividing $|G|.$ Write $|G|=p^k m$ with $k \ge 1$ and $p \nmid m.$

First Sylow Theorem.There is a subgroup $H\subseteq G$ of order $p^k.$ $H$ is called aSylow $p$-subgroup.

Second Sylow Theorem.Any two Sylow $p$-subgroups are conjugate: if $H$ and $K$ are Sylow $p$-subgroups, there is an element $g \in G$ such that $g^{-1}Hg = K.$

Third Sylow Theorem.Let $n_p$ be the number of Sylow $p$-subgroups. Then

$n_p|m$

$n_p \equiv 1 \pmod p$

$n_p = |G|/|N_G(H)|,$ where $H$ is any Sylow $p$-subgroup and $N_G(H)$ denotes the

normalizerof $H,$ the largest subgroup of $G$ in which $H$ is normal.

## Examples and Applications

Identify the Sylow subgroups of $S_4.$

$|S_4|=24=2^3 \cdot 3.$

The Sylow $2$-subgroups have order $8.$ There is a natural example of an order-8 subgroup in $S_4,$ namely the

dihedral group$D_4.$ Labeling the vertices of a square $1,2,3,4,$ the dihedral group is the set of symmetries of the square. It is generated by a $90$-degree rotation, which is a $4$-cycle, and a flip, which is a double transposition. There are four flips and four rotations (including the identity).The dihedral group is not normal inside $S_4;$ labeling the vertices differently leads to different copies. For instance, one copy is generated by $(13)(24)$ and $(1234),$ but another copy is generated by $(12)(34)$ and $(1324);$ and a third copy is generated by $(14)(23)$ and $(1243).$ These are all conjugate, as predicted by the second Sylow theorem; and $n_2=3$ meets the criteria of the third Sylow theorem: $n_2 \equiv 1 \pmod 2$ and $n_2 | 3.$

The Sylow $3$-subgroups are generated by the $3$-cycles, like $(123).$ There are $8$ such $3$-cycles, hence four subgroups (each subgroup contains two of them). So $n_3=4.$ This agrees with the third Sylow theorem, since $n_3|8$ and $n_3 \equiv 1 \pmod 3.$ As for $N_{S_4}(H),$ the third Sylow theorem predicts that this is a subgroup of order $|S_4|/n_3 = 24/4 = 6.$ Let $H = \langle (123) \rangle$ be one of the Sylow $3$-subgroups, then $N_G(H)$ is a copy of $S_3$ inside $S_4,$ consisting of all the permutations that fix $4.$ $_\square$

Note that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup. So if $n_p=1,$ so that the Sylow $p$-subgroup is unique, then it must be normal. This is a common source of applications.

Show that there is no simple group of order $30.$

Suppose $G$ is a simple group of order $30.$ The third Sylow theorem for $p=3$ says that $n_3|10$ and $n_3 \equiv 1 \pmod 3.$ So $n_3 = 1$ or $10.$ If $n_3=1$ then the Sylow $3$-subgroup is normal, which is impossible since $G$ is simple. So $n_3=10$ and there are $10$ Sylow $3$-subgroups. They are distinct, so they intersect only in the identity element (the intersection is a subgroup, and by Lagrange's theorem its order is $1$ or $3,$ so it must be $1$). Any Sylow $3$-subgroup has order $3,$ and consists of the identity and two elements of order $3.$ So there are $20$ elements of order $3$ in $G.$

Now $n_5|6$ and $n_5 \equiv 1 \pmod 5,$ so $n_5 =1$ or $6.$ As before, $n_5$ must be $6,$ so there are $6$ Sylow $5$-subgroups with $5$ elements each, consisting of the identity and four elements of order $5.$ Any pair of subgroups intersects only in the identity, so there are a total of $24$ elements of order $5$ in $G.$ But this count, coupled with the previous paragraph, gives at least $20+24=44$ elements in a group of order $30,$ which is impossible. So there is no simple group of order $30.$ $_\square$

## Abelian Groups

The Sylow theorems give extensive information about the structure of abelian groups (groups where the operation is commutative). Every subgroup of an abelian group is normal, so $n_p=1$ for all primes $p$ dividing $|G|.$ So there is a unique Sylow $p$-subgroup for every such prime $p.$

Next, there is a useful general lemma about abelian groups:

Let $H,K$ be subgroups of an abelian group $G,$ and suppose $H \cap K = \{1\}.$ Then $HK = \{hk \colon h \in H, k \in K\}$ is a subgroup of $G,$ and it is isomorphic to the Cartesian product $H \times K.$

Checking that $HK$ is a group boils down to checking that it is closed under the group operation $(h_1k_1)(h_2k_2) = (h_1h_2)(k_1k_2) \in HK$ because $G$ is abelian. (Note that if $G$ is not abelian, $HK$ as defined above is not necessarily a subgroup. Many references define $HK$ to be the smallest subgroup containing all the elements $hk$ for $h\in H,k\in K.$)

There is a natural homomorphism $H \times K \to HK$ given by $(h,k) \mapsto hk.$ It is surjective by definition, and to see that it is injective, suppose $(h,k) \mapsto 1.$ Then $hk=1,$ so $h = k^{-1} \in K.$ But $H \cap K = \{1\},$ so $h=1$ and $k=1.$ So the kernel of the homomorphism is trivial. $_\square$

Note that if $H$ and $K$ are finite, with relatively prime orders, their intersection must be trivial by Lagrange's theorem. So the lemma applies to them.

Consider the effect of repeatedly applying the lemma to the Sylow $p$-subgroups of an abelian group $G.$ Suppose $|G| = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_k^{\alpha_k}$ is the prime factorization of $|G|.$ Let $H_i$ be the unique Sylow $p_i$-subgroup. Then the lemma shows that $H_1H_2 \cong H_1 \times H_2.$ So $H_1H_2$ has order $p_1^{\alpha_1}p_2^{\alpha_2},$ and the lemma applies to the groups $H_1H_2$ and $H_3.$ So $H_1H_2H_3 \cong (H_1H_2) \times H_3 \cong H_1 \times H_2 \times H_3.$

Proceeding in the same way gives the following result:

A finite abelian group $G$ is isomorphic to the Cartesian product of its Sylow subgroups.

This is a key component in the structure theorem for finite abelian groups.