Every p-Group is Solvable

# 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.