Every p-Group is Solvable
 Proposition 1: Let $G$ be a $p$-group, where $p$ is a prime. Then $G$ is solvable.
Recall that a $p$-group is a group of order $p^k$ where $p$ is a prime and $k \geq 1$.
• Proof: Let $G$ be a $p$ group. Then $|G| = p^k$ for some $k \geq 1$. We construct a chain of subgroups that satisfy the definition for $G$ to be solvable.
• Let $G_0 = \{ e \}$ be the trivial group.
• Since $p \mid p^k = |G|$, we have by Burnside's Theorem for p-Groups that $Z(G)$ is nontrivial. Let $G_1 = Z(G)$. Then trivially, $G_0 = \{ e \}$ is a normal subgroup of $G_1 = Z(G)$, and $G_1/G_0 = Z(G)/\{ e \} \cong Z(G)$ is abelian.