The greatest enemy of knowledge is not ignorance, it is the illusion of knowledge, Daniel Boorstin.
Recall that a p-group is a group of order p^{k} where p is a prime and k ≥ 1.
Burnside’s Theorem for p-Groups. The center Z(G) of any group G of order p^{k}, with p prime is non-trivial.
Proof.
By definition, if z ∈ Z(G), then gzg^{-1} = z ∀ g ∈ G, so [z] is a trivial conjugacy class. Since conjugation is a group action of G on itself, and thus induces an equivalence relation on G, the conjugacy classes partition G.
The class equation for G is: $p^k = |Z(G)| + \sum_{g∈G,[g]non-trivial}|G : C_G(g)|$ where the sum of the righthand side of the equation runs over each element g with a non-trivial conjugacy class [g] = |G : C_{G}(g)| > 1.
|G| = p^{k} = [By Lagrange’s Theorem] |G : C_{G}(g)| |C_{G}(g)| ⇒ |G : C_{G}(g)| | p^{k}, but also |G : C_{G}(g)| > 1 ⇒ p | |G : C_{G}(g)|
Therefore, if [g] is a non-trivial conjugacy class we have that p | |G : C_{G}(g)| ⇒ p | $\sum_{g∈G,[g]non-trivial}|G : C_G(g)|$ and obviously p | p^{k} ⇒ $p \mid Z(G)$ ⇒ Z(G)≥ p ⇒ Z(G) is non-trivial ∎
Theorem. A group of order p, where p is prime, is solvable.
Proof.
Let G be a group of order p. Then, the only subgroup of G is the trivial group. The composition series for G is {id} ⊆ G which is a normal series, and its only factor is G/{id} = G ≋ [Every group of prime order is cyclic] ℤ/ℤ_{p} which is Abelian.
Theorem. If |G| = p^{n} where p is a prime number, then G is solvable. In other words, every p-group where p is a prime is solvable
Proof.
By induction on n.
n = 1 ⇒ |G| = p, p prime ⇒ [Previous Theorem. A group of order p, where p is prime, is solvable.] G is a cyclic group of prime order, thus it is solvable.
Let the statement holds for all k ≤ n.
Let’s prove it that it holds for n + 1. [Burnside’s Theorem for p-Groups. The center Z(G) of any group G of order p^{k}, with p prime is non-trivial.] Z(G) ≠ {e}. Futhermore, Z(G) ◁ G (∀g∈G, h∈Z(G), ghg^{-1} = h by definition of Z(G)) and Z(G) is Abelian ⇒ [Abelian groups are solvable] Z(G) is trivially solvable, {e} ◁ Z(G).
G/Z(G) is again a p-group or trivial (Before we have demonstrated that p | Z(G)).
Theorem. If |G| = p^{k}q^{s} where p and q are prime numbers, k, s ∈ ℕ, and 1 mod p ≠ q^{t} for t = 1, 2, ···, s, then G is solvable.
Proof
Recall: Let p be a prime number that divides |G|. Let k be the biggest natural number such that p^{k} divides |G|. All the subgroups of G with order p^{k} are called p-Sylow subgroups of G. We denote their set to be Syl_{p}G = {P : P ≤ G and |P|=p^{k}} and #Syl_{p}G = N_{p} = |Syl_{p}G|
By 3rd Sylow Theorem, N_{p} = |Syl_{p}G| divides the order of G, N_{p} | p^{k}q^{s} and N_{p} ≡ 1 mod p. By assumption, 1 mod p ≠ q^{t} for t = 1, 2, ···, s ⇒ N_{p} = |Syl_{p}G| = {P : P ≤ G and |P|=p^{k}} = 1. Let P be the only p-Sylow subgroup of G ⇒ [If G has only one p-Sylow subgroup H, then H is normal. This is because every two p-Sylow subgroups are conjugate, so if N_{p} = 1 ⇒ the p-Sylow subgroup is conjugate only to itself, i.e., it is normal] P ◁ G and by definition |P| = p^{k} ⇒ [Previous Theorem. Every p-group where p is a prime is solvable] P is solvable.
Besides, [G : P] = q^{s}, q prime ⇒ G/P is a q-group ⇒ Since every p-group where p is a prime is solvable, then G/P is solvable. P and G/P are solvable ⇒ [If G is a group, H ◁ G such that H is solvable and G/H is solvable ⇒ G is solvable] P is solvable.
In particular, if |G| = p·q where p and q are distinct prime numbers, then G is solvable (previous theorem for k = s = 1).
Exercise. Any group of order 10 is solvable.
|G| = 10 ⇒ By Cauchy’s theorem, G has a subgroup of order 5. Since this subgroup has index 2, it is automatically normal, say H ◁ G. Therefore, we construct the tower: {id} ⊆ H ⊆ G. It is a normal tower. Futhermore, H/{id} = H ≋ ℤ/5ℤ, Abelian; and G/H ≋ ℤ/2ℤ, Abelian, Hence G is solvable.