Show that a group of order pn, where p is prime, is solvable.
Show that a group of order pn, where p is prime, is solvable.
Let G be a group of order p^n, where p is a prime number.
We will prove by induction on n that G is solvable. The base case is n = 1, where G is a group of order p. Any group of prime order is necessarily cyclic and therefore abelian, so it is solvable.
Now suppose that G is a group of order p^n for some n > 1. By the class equation, there exists an element g ∈ G whose conjugacy class has size divisible by p. In particular, the centralizer C_G(g) of g in G has order at least p, and so the quotient group G/C_G(g) has order p^{n-1}.
By the induction hypothesis, G/C_G(g) is solvable, so there exists a chain of subgroups
{e} = N_0 ⊲ N_1 ⊲ ⋯ ⊲ N_{m-1} ⊲ N_m = G/C_G(g)
where each quotient group N_{i+1}/N_i is abelian. Letting H_i be the inverse image of N_i under the natural projection G → G/C_G(g), we obtain a chain of subgroups
{e} = H_0 ⊲ H_1 ⊲ ⋯ ⊲ H_{m-1} ⊲ H_m = G
where each quotient group H_{i+1}/H_i is isomorphic to (N_{i+1}/N_i) × C_G(g), which is also abelian. Thus, G is solvable, completing the induction.
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