Groupprops, The Group Properties Wiki (pre-alpha)

From Groupprops

Contents 1 Statement 2 Related facts 2.1 Other parts of Sylow's theorem 2.2 Stronger forms of existence 2.3 Analogues for Hall subgroups 3 Facts used 4 Proof 4.1 Proof by action on subsets 4.2 Proof by conjugation action

Statement

Let G be a finite group and p be a prime number. Then, there exists a p-Sylow subgroup P of G: a subgroup whose order is a power of p and whose index is relatively prime to p.

Note that when p does not divide the order of G, the p-Sylow subgroup is trivial, so the statement gives interesting information only when p divides the order of G. This statement is often viewed as a part of a more general statement called Sylow's theorem.

Related facts

Other parts of Sylow's theorem

• Sylow implies order-dominating: Given any p-Sylow subgroup and any p-subgroup, the p-Sylow subgroup contains a conjugate of the p-subgroup.
• Sylow implies order-conjugate: All p-Sylow subgroups are conjugate. This follows directly from the previous part.
• Congruence condition on Sylow numbers: The number of p-Sylow subgroups is congruent to 1 modulo p.

Stronger forms of existence

• Every Sylow subgroup intersects the center nontrivially or is contained in a centralizer: Every Sylow subgroup of a group either has a nontrivial intersection with the center of the group, or it is contained in the centralizer of some non-central element.

Analogues for Hall subgroups

The analogous statement does not hold for Hall subgroups. A Hall subgroup is a subgroup whose order and index are relatively prime. A π-Hall subgroup is a Hall subgroup such that the set of primes dividing its order is contained in π, and the set of primes dividing its index is disjoint from π.

• Hall subgroups need not exist: Given a set π of primes, there may not exist a π-Hall subgroup.
• Hall subgroups exist in finite solvable: If, however, the finite group is solvable, then it has π-Hall subgroups for all prime sets π.
• Hall's theorem on solvability: This states that a finite group is solvable if and only if it has π-Hall subgroups for every prime set π.

Facts used

1. Lucas's theorem
2. Fundamental theorem of group actions: There is a bijection between the coset space of the stabilizer of an element and the orbit of that element. In particular, the size of the orbit equals the index of the stabilizer.
3. Lagrange's theorem
4. Class equation of a group
5. Cauchy's theorem for Abelian groups
6. Central implies normal: Any subgroup inside the center is normal.

Proof

Proof by action on subsets

Given: A finite group G of order n = p^rm where p is prime, r is a nonnegative integer, and p does not divide m.

To prove: G has a subgroup of order p^r.

Proof: Consider the action of G by left multiplication on the set W of subsets of G of size p^r. Here are some observations regarding this action:

1. This action is well-defined: First, note that a group acts on the set of all its subsets by left multiplication. Further, since the left multiplication maps are bijective, they preserve the sizes of subsets. In particular, any subset of size p^r gets mapped to a subset of size p^r. Thus, we can restrict the action to the set W of subsets of size p^r.
2. The size of W is relatively prime to p: The size of W is a binomial coefficient, and an appeal to Lucas's theorem (fact (1)) reveals that its value is relatively prime to p.
3. The action of G on W has at least one orbit, say whose size is relatively prime to p: The size of W equals the sum of sizes of the orbits. Since the total is relatively prime to p, one of the numbers has to be relatively prime to p.
4. Let S be a member of the orbit and H be the stabilizer of S in G. Then, the index of S is relatively prime to p: By fact (2), the size of the orbit equals the index of the stabilizer of S in G.
5. The index of H is at least m: Notice that the union of , is the whole group G. Since each gS has size p^r, there are at least n / p^r = m translates of S in G. Thus, has size at least m.
6. The index of H is exactly m: By fact (3) (Lagrange's theorem), the index of H in G is a divisor of the order of G. By step (4), this index is relatively prime to p. Hence, the index of H in G divides m. Combining this with step (5), we obtain that the index of H in G is exactly equal to m.
7. The order of H is p^r: This follows from fact (3) (Lagrange's theorem) and the previous step.

Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory (although a purely algebraic proof also exists).

Proof by conjugation action

Given: A finite group G of order n = p^rm, where p is prime, r is a nonnegative integer, and p does not divide m.

To prove: G has a subgroup of order p^r.

Proof: We prove the claim by induction on the order of G. Specifically, we assume that the result is true for all groups of order smaller than the order of G.

For the case r = 0, the proof is direct since the trivial group is a subgroup of order p^r. Thus, we assume r > 0. In particular, p divides the order of G.

Consider the class equation of G (fact (4)):

where are the conjugacy classes of non-central elements and g_i is an element of c_i for each i.

We consider two cases:

1. Case that p divides the order of Z(G):
   1. There exists a normal subgroup of order p in G: Since Z(G) is Abelian, fact (5) yields that it has a subgroup H of order p. Since H is in the center, H is normal in G (by fact (6)). Thus, H is a normal subgroup of G of order p.
   2. G / H has a p-Sylow subgroup, in particular, a subgroup Q of order p^{r − 1}: Since H has order p, G / H has order p^rm / p = p^{r − 1}m. This is strictly smaller than the order of G, so induction applies and we get a p-Sylow subgroup, whose order is p^{r − 1}.
   3. Let be the quotient map. Then α ^{− 1}(Q) is a p-Sylow subgroup of G: Indeed, the order of α ^{− 1}(Q) is the product of the order of H and the order of Q, which is therefore .
2. Case that p does not divide the order of Z(G):
   1. There exists i such that p does not divide the index k of C_G(g_i) in G: Since p divides the order of G, p cannot divide the index of every C_G(g_i), otherwise the class equation would yield that p divides the order of Z(G).
   2. C_G(g_i) is a proper subgroup of G whose order is a multiple of p^r: Since g_i is non-central, C_G(g_i) is proper in G. Further, since | G:C_G(g_i) | = k is relatively prime to p, Lagrange's theorem (fact (3)) yields that the order of C_G(g_i) is p^rm / k, which is a multiple of p^r.
   3. C_G(g_i) contains a subgroup of order p^r: This follows by the induction hypothesis.
   4. G contains a subgroup of order p^r: A subgroup of order p^r in C_G(g_i) is also a subgroup of order p^r in G.