Every p-Group is Solvable
 Table of Contents

# 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.
• Note that $G_1 = Z(G)$ is normal in $G$ (since for all $g \in G$ and for all $h \in Z(G)$ we have that $ghg^{-1} = h$ by definition of $Z(G)$, so that $G_1 = Z(G)$ is normal in $G$). So the quotient group $G/G_1$ is defined and furthermore $|G/G_1| = |G|/|G_1| = p^k/p = p^{k-1}$. If $p^{k-1} = 1 then [[$ G_1 = G$, and$\{ e \} = G_0 \leq G_1 = G$satisfies the properties for$G$being solvable. • If$|G/G_1| = p^{k-1} \neq 1$, then since$p$is a prime and$p \mid |G/G_1|$we again have by Burnside's Theorem that$Z(G/G_1)$is nontrivial. Let$G_2$be a subgroup of$G$containing$G_1$that is isomorphic to$Z(G/G_1)$. • We continue this process until it terminates at some$G_n = G$(which is must since$G$is a finite group) to obtain a chain of successive subgroups$\{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G$with$G_i$being a normal subgroup of$G_{i+1}$for each$0 \leq i \leq n-1$, and$G_{i+1}/G_i$being abelian for all$0 \leq i \leq n - 1$. • Thus$G$is a solvable group.$\blacksquare$# Example 1 It can be shown that for every prime$p$there exists a nonabelian group of order$p^3\$. By the proposition above, each of these nonabelian groups must be solvable.

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