Composition Factor Criterion for a Finite Group to be Solvable

# Composition Factor Criterion for a Finite Group to be Solvable

Recall from the Solvable Groups page that a group $G$ is said to be solvable if there exists a (finite) chain of successive subgroups:

(1)\begin{align} \quad \{ 1 \} = G_0 \leq G_1 \leq ... \leq G_n = G \end{align}

With the following properties:

**1)**$G_i$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq n -1$.

**2)**$G_{i+1}/G_i$ is an abelian group for all $0 \leq i \leq n-1$.

We noted that every abelian group is a solvable group.

The following theorem gives us a criterion for determining whether or not a finite group is solvable based on composition factors.

Theorem 1: Let $G$ be a finite group. Then $G$ is solvable if and only if there exists a (finite) chain of successive subgroups $\{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G$ such that:1) $G_i$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq n - 1$.2) $G_{i+1}/G_i$ is cyclic and has prime order for all $0 \leq i \leq n - 1$. |

**Proof:**$\Rightarrow$ Suppose that $G$ is solvable and let:

\begin{align} \quad \{ e \} = H_0 \leq H_1 \leq ... \leq H_n = G \quad (*) \end{align}

- be a (finite) chain of successive subgroups such that $H_i$ is a normal subgroup of $H_{i+1}$ for all $0 \leq i \leq n - 1$ and where $H_{i+1}/H_i$ is abelian for all $0 \leq i \leq n - 1$.

- Without loss of generality, we can assume that each composition factor $H_{i+1}/H_i$ is nontrivial, for if $H_{i+1}/H_i = \{ 1 \}$ then $H_i = H_{i+1}$ and we can just delete $H_i$ from the chain above - which still satisfies the properties defining $G$ to be solvable.

- Suppose that the composition factor $H_{i+1}/H_i$ is not cyclic of prime order for some $i$. Since $H_{i+1}/H_i$ is nontrivial, $|H_{i+1}/H_i| > 1$ and so let $p$ be a prime such that $p \mid |H_{i+1}/H_i|$.

- Since $p$ is a prime and $p \mid |H_{i+1}/H_i|$, by Cauchy's Theorem for Groups there exists an element $h^*H_i \in H_{i+1}/H_i$ with $\mathrm{ord}(h^*H_i) = p$.

- Now note that subgroups of $H_{i+1}/H_i$ correspond to subgroups of $H_{i+1}$ that contain $H_i$. Let $q : H_{i+1} \to H_{i+1}/H_i$ and let $H = q^{-1}(aH_i)$. Then:

\begin{align} \quad H_i \leq H \leq H_{i+1} \end{align}

- Note that $H$ is a normal subgroup of $H_{i+1}$ since $\langle aH_i \rangle$ is a normal subgroup of the abelian group $H_{i+1}/H_i$ (since $\langle a H_i \rangle$ is cyclic, and thus abelian). Furthermore, $H_i$ is normal in $H$ since $H_i$ is normal in $H_{i+1}$ (i.e., since $gH_ig^{-1} = H_i$ for all $g \in H_{i+1}$ we see that $gH_ig^{-1} = H_i$ for all $g \in H$).

- Let $f : H_{i+1}/H_i \to H_{i+1}/H$ be defined for all $h \in H_{i+1}/H_i$ by $f(hH_i) = hH$. Since $H_{i+1}/H_i$ is abelian, so is $f(H_{i+1}/H_i) = H_{i+1}/H$.

- Furthermore, we have that $H/H_i \cong \langle h^* H_i \rangle$, i.e., $H/H_i$ is cyclic of prime order, and is also thus abelian. We the insert $H$ into the series at $(*)$ and continue in this process until all composition factors are cyclic of prime order.

- $\Leftarrow$ Suppose that there exists a (finite) chain of successive subgroups satisfying (1) and (2). Since each composition factor is cyclic and of prime order we have that each composition factor is abelian (since every cyclic group is abelian). Thus $G$ is a solvable group. $\blacksquare$