The Sylow Theorems and Groups of Small Order

Let G be finite and let p divide |G|. Write |G|=p^a m, where p does not divide m. A Sylow p-subgroup has order p^a, the largest power of p dividing |G|. Sylow's theorems say these subgroups exist and, for this fixed prime, are all conjugate. Their number n_p divides m and is 1 modulo p. Thus n_p leaves remainder 1 when divided by p.

A Sylow p-subgroup is normal exactly when n_p=1. Uniqueness makes every conjugate equal to it. Conversely, normality fixes it under conjugation, and Sylow conjugacy leaves no other one. In an abelian group every subgroup is normal, so each Sylow count is 1. A simple group is nontrivial and has no nontrivial proper normal subgroup. A normal Sylow subgroup rules out simplicity when its order lies strictly between 1 and |G|.

Suppose all Sylow subgroups are normal and |G| is composite. With two distinct prime divisors, a Sylow subgroup is nontrivial and proper, so G is not simple. If |G|=p^a with a>=2, its Sylow subgroup is G itself. Instead, the class equation gives a nonidentity central element z. Its order is p^b. Then z^(p^(b-1)) has order p and generates a central, hence normal, subgroup. Since 1<p<p^a, this subgroup rules out simplicity too.

Distinct subgroups of prime order p share only the identity. A shared nonidentity element would generate both. If p divides |G| but p^2 does not, these are the Sylow p-subgroups, and exactly n_p(p-1) elements have order p. For a distinct prime q with q dividing |G| but q^2 not dividing |G|, the count is n_q(q-1). The two element sets are disjoint, so their sum is at most |G|-1.

A complement count can prove uniqueness, but it must establish containment first. Suppose the complement of all order-p elements has exactly q^b elements, where p and q are different primes. Every Sylow q-subgroup contains no element of order p, since its element orders divide a power of q. If each Sylow q-subgroup also has q^b elements, it equals that complement. They are therefore all the same subgroup, which is normal.

For normal subgroups P and Q of coprime orders, their intersection is {e}: its order divides both orders. For x in P and y in Q, xyx^-1y^-1 belongs to P because yx^-1y^-1 is in P. It belongs to Q because xyx^-1 is in Q. The intersection is trivial, so xyx^-1y^-1=e and xy=yx. Also, xy=x'y' gives x'^-1x=y'y^-1 in the intersection. Hence x=x' and y=y': all product expressions are unique.

Suppose |G|=pq for distinct primes and both Sylow subgroups are normal. Prime-order groups are cyclic, so write P=<a> and Q=<b>, with a and b of orders p and q. These subgroups commute and intersect in {e}. Their pq distinct products fill G. If (ab)^k=e, then a^k=b^-k is in the intersection. Thus p and q both divide k. The least positive such k is pq, so ab generates G and G is cyclic.

Let |G|=2p, where p is an odd prime. The Sylow p count divides 2 and is 1 modulo p, so it is 1. Choose a generator a of this normal subgroup P and an element b of order 2 from a Sylow 2-subgroup. Since b is outside P, the two cosets are P and Pb. Every element is uniquely a^i b^j, where 0<=i<p and j is 0 or 1. It remains to determine how b conjugates a.

Normality gives bab^-1=a^r for a nonzero residue r modulo p. Conjugating twice sends a to a^(r^2). Since b^2=e, this equals a, so r^2=1 modulo p. The prime p divides (r-1)(r+1), hence divides one factor. Thus r is 1 or -1 modulo p. The relation ba=a^r b reduces every product to one of the forms a^i b^j, so these two possibilities determine all multiplication.

For r=1 the generators commute, and ab has order 2p, giving the cyclic group. For r=-1 the multiplication is that of rotations and reflections of a regular p-gon: take a as a basic rotation and b as a reflection. Their 2p distinct forms match all group elements and all products. Both groups exist. One is abelian and the other is not, so they are not isomorphic. These are exactly the two groups of order 2p.

In the inversion case, ba^i=a^-i b. Therefore (a^i b)^2=a^i a^-i b^2=e. The p distinct elements a^i b are outside P, so none is the identity; each has order 2. Every nonidentity element of P has the odd prime order p. Thus exactly p elements have order 2. For p=3, the rotations and reflections act on the triangle's vertices as all permutations of three letters.

Recall orbit-stabiliser for a group action. The normaliser N_G(P) is the subgroup of elements g satisfying gPg^-1=P. It is the stabiliser of P under conjugation on the Sylow p-subgroups. Sylow conjugacy makes this action transitive, with orbit size n_p. Hence n_p times |N_G(P)| equals |G|. Divide |G| by n_p to find the normaliser's order. The normaliser equals G exactly when P is normal.