# 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 $$g \in A \setminus H$$. Let $$t$$ be the order of $$g$$. Consider the group $$A' = H \langle g \rangle$$. Since $$A$$ is abelian, we have $$H \langle g \rangle = \langle g \rangle H$$ thus $$A'$$ is a group. But since it strictly contains $$H$$, we have $$A' = A$$ by maximality of $$H$$.

Now $$H \langle g \rangle$$ contains $$|H|t / d$$ elements where $$d = |H \cap \langle g \rangle|$$. 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 $$h_i \gt 1$$ 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 $$P = \langle g \rangle$$ 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 $$1 \ne h\in H$$ then the order of $$h$$ is $$p^\mu$$ for some $$\mu \gt 0$$. Then $$h^{p^{\mu - 1}}$$ 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 $$k = 1 \bmod p$$ 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 $$A\triangleleft N$$. 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 $$D_i = g^{-1}_i A g_i \cap N$$. 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 $$p \ne 1 \pmod{q}$$ and $$q \ne 1 \pmod{p}$$ is abelian.

Proof: We have already shown this for $$p = q$$ so assume $$(p,q)=1$$. Let $$P = \langle a \rangle$$ 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 $$q \ne 1 \bmod p$$. 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 $$Q = \langle b \rangle$$ 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 $$a^\alpha b^\beta = b^\beta a^\alpha$$ 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 $$\langle a \rangle$$ 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 $$p^\mu$$ for any $$0 \lt \mu \lt m$$.

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 $$m \gt 2$$. 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 $$p^{\mu - 1}$$ which has the form $$A/P$$ for some normal subgroup $$A$$ in $$G$$ with order $$p^\mu$$.

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

$G \triangleright A_1 \triangleright A_2 \triangleright ... \triangleright A_{m-1} \triangleright \{1\}$

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 $$k = 1 \bmod 5$$ 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 $$n\ge 5$$:

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 $$|H| \ge 1 + 6\cdot 4 = 25$$ 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 $$\langle (1 2 3 4 5) \rangle$$ and $$\langle (1 3 2 4 5) \rangle$$ are distinct Sylow 5-subgroups.

Theorem:$$A_n$$ is simple for all $$n\ge 5$$.

TODO: proof

Ben Lynn blynn@cs.stanford.edu 💡