Group Theory - Sylow Groups

Lemma: Let $A$ be an abelian group. If $p$ is a prime factor of $∣ A ∣$ then $A$ contains at least one element of order $p$

Proof: The lemma is trivial when $∣ A ∣ = p$ , which we shall use to start an induction. Assume $∣ A ∣$ is composite. Then $A$ contains a proper subgroup. Choose a proper subgroup $H$ of maximum order. If $p ∣ ∣ H ∣$ then by induction $H$ contains an element of order $p$ , so assume $(∣ H ∣, p) = 1$ . Then take some element $g \in A ∖ H$ . Let $t$ be the order of $g$ . Consider the group $A' = H ⟨ g ⟩$ . Since $A$ is abelian, we have $H ⟨ g ⟩ = ⟨ g ⟩ H$ thus $A'$ is a group. But since it strictly contains $H$ , we have $A' = A$ by maximality of $H$ .

Now $H ⟨ g ⟩$ contains $∣ H ∣ t / d$ elements where $d = ∣ H \cap ⟨ g ⟩ ∣$ . Thus $∣ A ∣ d = ∣ H ∣ t$ . Since the $p$ divides the left-hand side, and $(∣ H ∣, p) = 1$ , we must have $p ∣ t$ , and $g^{t / p}$ is an element of order $p$ .

If the order of a group $G$ is divisible by $p^{m}$ but by no higher power of $p$ for some prime $p$ then any subgroup of $G$ of order $p^{m}$ is called a Sylow group corresponding to $p$ .

Theorem: Every group $G$ possesses at least one Sylow group corresponding to each prime factor of $∣ G ∣$ .

Proof: The theorem is immediate when $∣ G ∣ = 2$ , which we shall use to start an induction. Write $∣ G ∣ = p^{m} r$ where $(r, p) = 1$ . Decompose $G$ into classes of conjugate elements, and pick elements $a_{1}, . . ., a_{k}$ from each class. Recall if $h_{i}$ denotes the size of the class containing $a_{i}$ we have $∣ G ∣ = h_{1} + . . . + h_{k}$ . Also recall the normalizer $N_{i}$ of $a_{i}$ satisfies $∣ N_{i} ∣ = ∣ G ∣ / h_{i}$ . We have two cases:

Case 1: Suppose there exists $h_{i}$ with $h_{i} > 1$ and $(h_{i}, p) = 1$ . Then $∣ N_{i} ∣$ is less than $∣ G ∣$ and divisible by $p^{m}$ . By inductive hypthoesis, $N_{i}$ possesses a subgroup of order $p^{m}$ which is the Sylow group corresponding to $p$ .

Case 2: For all $i$ , we have $h_{i} = 1$ or $p ∣ h_{i}$ . We have $h_{i} = 1$ for self-conjugate elements, and we must have at least one of these since $1$ is self-conjugate. Let $z$ be the number of self-conjugate elements. Then $p^{m} r = z + x p$ for some integer $x$ , hence $p ∣ z$ . Thus the order of the center is divisible by $p$ . Since it is abelian, by the lemma it contains at least one element $g$ that commutes with all elements and has order $p$ . Then $P = ⟨ g ⟩$ is a normal subgroup of $G$ and $G / P$ has order $p^{m - 1} r$ . By the inductive hypothesis $G / P$ contains a Sylow group of order $p^{m - 1}$ , which we write $H / P$ where $H$ is a subgroup of $G$ . Then $p^{m - 1} = ∣ H ∣ / p$ , thus $∣ H ∣ = p^{m}$ and $H$ is a Sylow group of $G$ corresponding to $p$ .

Theorem: [Cauchy] Let $G$ be a group. If $p$ is a prime factor of $∣ G ∣$ then $G$ contains at least one element of order $p$

Proof: Let $H$ be a Sylow group of $G$ of order $p^{m}$ . If $1 \neq h \in H$ then the order of $h$ is $p^{μ}$ for some $μ > 0$ . Then $h^{p^{μ - 1}}$ has order $p$ .

All subgroups conjugate to a Sylow group are themselves Sylow groups. It turns out the converse is true.

Theorem: All Sylow groups belonging to the same prime are conjugates.

Proof: Let $A, B$ be subgroups of $G$ of order $p^{m}$ . Recall we can decompose $G$ relative to $A$ and $B$ : $G = A g_{1} B \cup . . . \cup A g_{r} B$ and $∣ G ∣ = ∣ A ∣ B ∣ / d_{1} + . . . + ∣ A ∣ ∣ B ∣ / d_{r}$ where $d_{i}$ is the size of $D_{i} = g_{i}^{- 1} A g_{i} \cap B$ . We have $∣ A ∣ = ∣ B ∣ = p^{m}$ and $∣ G ∣ = p^{m} r$ where $(r, p) = 1$ . Thus dividing by $p^{m}$ gives $r = \frac{p^{m}}{d_{1}} + . . . \frac{p^{m}}{d_{r}}$ Now $D_{i}$ is a subgroup of $B$ , hence $d_{i}$ is some nonnegative power of $p$ and is at most $p^{m}$ . Since $(r, p) = 1$ , we must have $p^{m} / d_{l} = 1$ for some $l$ , in other words $d_{l} = p^{m}$ . Then $D_{l}$ has the same order as $B$ and is contained in $B$ , thus $D_{l} = B$ and similarly $D_{l} = g_{l}^{- 1} A g_{l}$ . Hence $B = g_{l}^{- 1} A g_{l}$ implying that $A, B$ are conjugate.

Corollary: A Sylow group is unique if and only if it is a normal subgroup.

Theorem: If there are exactly $k$ Sylow groups of a group $G$ corresponding to a prime $p$ then $k = 1 \mod p$ and $k$ divides $∣ G ∣$ .

Proof: We know that the number of distinct Sylow groups is equal to the number $k$ of distinct conjugates. Let $A$ be some Sylow group corresponding to $p$ and let $N$ be the normalizer of $A$ . Recall $∣ G ∣ = ∣ N ∣ k$ thus $k$ divides $∣ G ∣$ .

Every $a \in A$ satisfies $a^{- 1} A a = A$ thus $a \in N$ , Hence $A ◃ N$ . Thus $∣ N ∣ = p^{m} n'$ where $(n', p) = 1$ .

Decompose $G$ as the disjoint sets $G = A g_{1} N \cup . . . \cup A g_{r} N$ Then $∣ G ∣ = \frac{∣ N ∣ p^{m}}{d_{1}} + . . . + \frac{∣ N ∣ p^{m}}{d_{r}}$ where $d_{i}$ is the order of the group $D_{i} = g_{i}^{- 1} A g_{i} \cap N$ . Without loss of generality assume $g_{1} = 1$ , hence $A g_{1} N = A N = N$ . Now dividing by $n$ gives $k = 1 + \frac{p^{m}}{d_{2}} + . . . + \frac{p^{m}}{d_{r}}$ Now suppose $d_{i} = p^{m}$ for some $i$ . Then $D_{i} = g_{i}^{- 1} A g_{i}$ , implying $g_{i}^{- 1} A g_{i} \subset N$ . Now $N$ possesses a Sylow group of order $p^{m}$ , and we have already found two: $A, g_{i}^{- 1} A g_{i}$ . But $A$ is normal in $N$ thus must be the unique Sylow group, hence $A = g_{i}^{- 1} A g_{i}$ . Since $N$ is the normalizer of $A$ we must have $g_{i} \in N$ and hence $A g_{i} N = A N = N$ , which is impossible unless $i = 1$ .

Thus all terms in the above summation are divisble by $p$ except for the first term which is equal to one.

Theorem: Any group $G$ of order $p q$ for primes $p, q$ satisfying $p \neq 1 (\mod q)$ and $q \neq 1 (\mod p)$ is abelian.

Proof: We have already shown this for $p = q$ so assume $(p, q) = 1$ . Let $P = ⟨ a ⟩$ be a Sylow group of $G$ corresponding to $p$ . The number of such subgroups is a divisor of $p q$ and also equal to $1$ modulo $p$ . Also $q \neq 1 \mod p$ . Then since the number of such subgroups cannot be equal to $p, q, p q$ , it must be equal to one. By the above corollary we have that $P$ is normal in $G$ of order $p$ . Similarly we can find a group $Q = ⟨ b ⟩$ normal in $G$ of order $q$ .

Then $P Q = Q P$ , which by the product theorem is a subgroup order $p q / ∣ P \cap Q ∣$ . But since $(p, q) = 1$ they only have the identity element in common thus $G = P Q$ . Also, recall these conditions also imply every element of $P$ commutes with every element of $Q$ . Then every element of $G$ has the form $a^{α} b^{β} = b^{β} a^{α}$ and is clearly abelian

A prime power group is a group whose order is a power of a prime. [It seems that nowadays they are referred to as $p$ -groups.] All Sylow groups are prime power groups. Recall that a group $G$ of order $p^{m}$ for a prime $p$ has at least one nontrivial self-conjugate element, thus we can find a self-conjugate element of order $p$ . Let $a$ be such an element. Then $x^{- 1} a x$ for any $x \in G$ , and $⟨ a ⟩$ is a normal subgroup of order $p$ . In general:

Theorem: A group of order $p^{m}$ for a prime $p$ contains at least one normal subgroup of order $p^{μ}$ for any $0 < μ < m$ .

Proof: The theorem is true for $m = 2$ because in this case the group is abelian. We shall use this case to base an induction.

Suppose $G$ is a group of order $p^{m}$ for $m > 2$ . Then let $P$ be a normal subgroup of $G$ of order $p$ . Then $G / P$ has order $p^{m - 1}$ which by inductive assumption has an invariant subgroup of order $p^{μ - 1}$ which has the form $A / P$ for some normal subgroup $A$ in $G$ with order $p^{μ}$ .

Corollary: All prime power groups are soluble.

Proof: A group $G$ of order $p^{m}$ has a normal subgroup $A_{1}$ of order $p^{m - 1}$ which in turn contains a normal subgroup of order $p^{m - 2}$ , and so on. Thus we can construct the composition series $G ▹ A_{1} ▹ A_{2} ▹ . . . ▹ A_{m - 1} ▹ {1}$

Example: There is no simple group of order 200. For let $G$ be a group with order 200. Then since $200 = 5^{2} \times 8$ , $G$ contains $k$ Sylow groups of order 25 where $k = 1 \mod 5$ and $k ∣ 200$ . Thus $k ∣ 8$ which is impossible unless $k = 1$ . Thus there exists a unique normal Sylow group of order 25, and hence the group is not stimple.

Example: There is no simple group of order 30. Suppose there is such a group. Then none of its Sylow groups are unique, implying it has $1 + 5 = 6$ Sylow groups of order 5, hence there are $6 \times 4 = 24$ elements of order $5$ , and similarly we must have $1 + 3 \times 3 = 10$ Sylow groups of order 3, thus the total number of elements is greater than 30, a contradiction.

We can now supply an alternative proof that $A_{n}$ is simple for $n \geq 5$ :

Proposition: If $∣ G ∣ = 60$ and $G$ has more than one Sylow 5-subgroup then $G$ is simple.

Proof: Suppose $∣ G ∣ = 60$ and contains more than one Sylow 5-subgroup, but there exists a proper normal subgroup. Then note we must have exactly $6$ Sylow 5-subgroups. Let $P$ be such a group. Then the normalizer of $P$ has order 10 since its index is $6$ .

If $5 ∣ ∣ H ∣$ then $H$ contains a Sylow $5$ -subgroup of $G$ and since $H$ is normal it contains all $6$ conjugates of this subgroup, hence $∣ H ∣ \geq 1 + 6 \cdot 4 = 25$ hence we must have $∣ H ∣ = 30$ . But by the previous example, $∣ G ∣$ must have a unique Sylow 5-subgroup, a contradiction, thus 5 does not divide $∣ H ∣$ .

If $∣ H ∣$ is 6 or 12 then $H$ has a normal Sylow subgroup of order 2,3, or 4, which is also normal in $G$ , and we may replace $H$ by this. Hence $∣ G / H ∣ = 30, 20$ or $15$ . Then by previous results, $G / H$ has a normal subgroup of order $5$ . Its preimage under the natural map is a normal subgroup whose order is a multiple of 5, which we have previously shown to be a contradiction.

Corollary: $A_{5}$ is simple.

Proof: The subgroups $⟨ (12345) ⟩$ and $⟨ (13245) ⟩$ are distinct Sylow 5-subgroups.

Theorem: $A_{n}$ is simple for all $n \geq 5$ .

TODO: proof