The Sylow Theorems

The Sylow theorems describe the p-subgroups of a finite group, where p is a prime number.

Definition. Let G be a group, and let p be a prime number. A Sylow p-subgroup is a maximal p-subgroup of G.

To be specific, if H is a Sylow p-subgroup of G and K is a p-subgroup of G such that $H \subset K$ , then $H = K$ .

Remark. Every p-subgroup of G is contained in a Sylow p-subgroup.

Suppose that G is finite, and let H be a p-subgroup. If H isn't properly contained in any other p-subgroup, it's a Sylow p-subgroup by definition. Otherwise, let $H'$ be a p-subgroup properly containing H.

Continuing in this way, I get a properly increasing chain of p-subgroups

$$H \propersubset H' \propersubset H'' \propersubset \cdots.$$

But G is finite, so this chain must be finite, and the process must stop with a Sylow p-subgroup.

If G is infinite, the result is still true, but requires the use of Zorn's lemma. I won't prove it here, since I'm only considering finite groups in this section.

Lemma. Let G be a group and let p be a prime number. Suppose that H is a normal p-subgroup of G and $G/H$ is a p-group. Then G is a p-group.

Proof. Let $g \in
   G$ . I must show that the order of g is a power of p.

Since $G/H$ is a p-group, the order of $gH$ is a power of p. Suppose that $(gH)^{p^n} = H$ , i.e. $g^{p^n}H = H$ . Thus, $g^{p^n} \in H$ .

Since H is a p-group, the order of $g^{p^n}$ is a power of p. Suppose that $(g^{p^n})^{p^m} = 1$ . Then $g^{p^{n+m}} = 1$ , and the order of g must be a divisor of $p^{n+m}$ . Therefore, the order of g is a power of p, and G is a p-group.

Lemma. Let G be a group, and let p be a prime number. G acts on the set of Sylow p-subgroups of G by conjugation.

Proof. The result is vacuously true if there are no Sylow p-subgroups. Otherwise, it suffices to show that that conjugate of a Sylow p-subgroup is a Sylow p-subgroup, since after that the axioms for an action are clearly satisfied.

Let H be a Sylow p-subgroup of G, and let $g \in G$ . I must show that $g H g^{-1}$ is a Sylow p-subgroup.

Since conjugation is an automorphism, it carries subgroups to subgroups. Therefore, $g H g^{-1}$ is a subgroup.

If $h \in H$ , then h has order $p^n$ for some $n \ge 0$ . Since conjugation is an automorphism, it preserves the orders of elements. Therefore, $g h g^{-1}$ also has order $p^n$ . It follows that $g H g^{-1}$ is a p-subgroup.

Finally, I must show that $g H g^{-1}$ is maximal. Suppose that K is a p-subgroup and $g H g^{-1} \subset K$ . Then $H \subset g^{-1} K g$ , and $g^{-1} K g$ is a p-subgroup because conjugation by $g^{-1}$ is an automorphism. Since H is maximal, $H = g^{-1} K g.$ Hence, $g H g^{-1} = K$ .

This proves that $g H g^{-1}$ is a Sylow p-subgroup, as I wished to show.

Remarks.

(a) The isotropy group of a Sylow p-subgroup H under the conjugation action is the normalizer $N(H)$ .

(b) If G has only one Sylow p-subgroup, it must be normal.

Lemma. Let G be a group, let p be a prime number, and let H be a Sylow p-subgroup. Then no nontrivial element of $N(H)/H$ has order which is a power of p.

Proof. Suppose on the contrary that $x H$ has order a power of p, where $x \in N(H)$ . I want to show that $x H$ is the identity.

If the order of $x H$ is $p^k$ , then $|\langle x H \rangle| = p^k$ . Then any element of $\langle x H \rangle$ has order dividing $p^k$ , so $\langle x H
   \rangle$ is a p-subgroup of $N(H)/H$ .

Subgroups of $N(H)/H$ are in one-to-one correspondence with subgroups of $N(H)$ which contain H. Thus, there is a subgroup K of $N(H)$ such that $K/H \approx \langle x H
   \rangle$ .

Now H and $\langle x H \rangle$ are p-groups, so K is a p-group, by an earlier lemma. Since K is a p-group and $H \subset K$ , maximality of H implies that $H = K$ .

Thus, $\langle x H \rangle = \{1\}$ , which is what I wanted to show.

Lemma. Let G be a group, let p be a prime number, let H be a Sylow p-subgroup, and let $g \in G$ . If the order of g is a power of p and $g H g^{-1} = H$ , then $g \in H$ .

Proof. If $g H
   g^{-1} = H$ , then $g \in N(H)$ . Since the order of g is a power of p, the same is true of the order of $gH \in N(H)/H$ . By the preceding lemma, $gH = H$ , and $g \in H$ .

Theorem. (Sylow) Let G be a finite group, and let p be a prime number.

(a) The number of Sylow p-subgroups is congruent to 1 mod p.

(b) Any two Sylow p-subgroups are conjugate.

(c) The number of Sylow p-subgroups divides $|G|$ .

Proof. Let H be a Sylow p-subgroup. Consider the set of conjugates of H

$$S = \{H = H_1, H_2, \ldots, H_k\}.$$

By an earlier result, all the H's are Sylow p-subgroups.

G acts on S by conjugation. Since $H <
   G$ , H also acts on S by conjugation. Consider the orbits under this action of H. The order of an orbit is the index $(H:H_s)$ of a corresponding isotropy group, and $(H:H_s) \mid |H|$ . Since the order of H is a power of p, it follows that the order of each orbit is a power of p.

Thus, the order of an orbit is divisible by p --- unless the orbit has order $p^0 = 1$ . In this case, $h H_i
   h^{-1} = H_i$ for some i and all $h \in H$ . Applying the preceding lemma, I find that $h \in H_i$ for all $h \in H$ , i.e. $H \subset H_i$ . Since $H_i$ is a Sylow p-subgroup, this implies that $H =
   H_i$ .

Thus, if an orbit consists of a single element, then it is $\{H\}$ . Conversely, if $h \in H$ , then $hHh^{-1} = H$ .

Thus, the orders of the orbits of S under conjugation by H are all divisible by p, except for $\{H\}$ . It follows that $|S|$ , which is the number of conjugates of H, is congruent to 1 mod p.

Next, let K be a Sylow p-subgroup, and suppose that K is not conjugate to H. As above, since G acts of S by conjugation and $K < G$ , K acts on S by conjugation. Again arguing as above, I find that the order of each orbit of S under conjugation by K is a power of p.

Could there be an orbit consisting of a single element? Again arguing as above, I conclude that if $\{H_i\}$ is a one-element orbit, then $H_i = K$ . But this is ruled out, because K was not a conjugate of H.

In this case, then, the orders of all the orbits of S under conjugation by K are divisible by p. It follows that $|S| = 0 \mod{p}$ , which contradicts the fact that $|S| = 1 \mod{p}$ established earlier.

This proves that every Sylow p-subgroup is a conjugate of H.

Finally, the number of conjugates of H is equal to the index $(G:N(H))$ , which divides $|G|$ . Therefore, the number of Sylow p-subgroups divides $|G|$ .

Corollary. (Sylow) Let G be a group of order $p^n m$ , where p is prime and $(p,m) = 1$ . Then every Sylow p-subgroup of G has order $p^n$ .

Proof. Let H be a Sylow p-subgroup of G. I have

$$(G:H) = (G:N(H))(N(H):H).$$

Since $(G:N(H))$ is the number of Sylow p-subgroups, it's congruent to 1 mod p by the preceding theorem. Therefore, $(G:N(H))$ is not divisible by p.

I claim that $(N(H):H)$ is not divisible by p. Indeed, if $p \mid (N(H):H)$ , then by Cauchy's theorem $N(H)/H$ has an element of order p. This contradicts an earlier lemma.

Since p does not divide $(G:N(H))$ or $(N(H):H)$ , it does not divide $(G:H)$ .

Now $|H| = p^k$ for some $k \le n$ , so

$$p^n m = |G| = (G:H)|H| = (G:H)p^k.$$

Then $p^{n-k}m = (G:H)$ , but $p
   \notdiv (G:H)$ . Therefore, $n - k = 0$ , and $n = k$ . This shows that $|H| = p^n$ .


Example. Let G be a group of order 15. Show that G is cyclic.

If H and K are subgroups of order 5, then $H \cap K$ is a subgroup of H, so $|H \cap K| \mid |H| = 5$ . Hence, $|H \cap K| = 1$ or $|H \cap K| = 5$ . In the second case, the subgroups are identical. In the case where the subgroups are distinct, they intersect in the identity only, and each contains 4 elements which are not in the other subgroup.

The number of Sylow subgroups of order 5 is congruent to 1 mod 5, so it could be 1, 6, 11, 16, .... None of 6, 11, or 16 divides 15, and if there were more than 16 Sylow-5 subgroups, the preceding discussion shows that they would contain more elements than there are in the group!

Consequently, there is a single Sylow-5 subgroup. Since there is a unique subgroup of order 5, it must be normal.

Arguing in the same way as with the Sylow-5 subgroups, I find that two Sylow-3 subgroups are either identical, or intersect only in the identity. Consequently, two distinct Sylow-3 subgroups each contains 2 elements which are not in the other.

The number of Sylow subgroups of order 3 is congruent to 1 mod 3, so it could be 1, 4, 7, .... Now 4 and 7 don't divide 15, and if there are more than 7 Sylow-3 subgroups, the preceding argument shows that there are at least $7\cdot 2 + 1 = 15$ elements in these subgroups. But I already have 4 elements of order 5 in the Sylow-5 subgroup, so this is impossible.

Consequently, there is a single Sylow-3 subgroup. Since there is a unique subgroup of order 3, it must be normal.

The Sylow-3 and Sylow-5 subgroups account for all the elements of order 3 or order 5, and there are $2 + 4
   + 1 = 7$ elements between them. This leaves $15 - 7 = 8$ elements unaccounted for. Since elements of G can have order 1, 3, 5, or 15, there must be elements of order 15, so G is cyclic.


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