# Solvable group

A solvable group is a type of group of particular interest, particularly in Galois theory.

A group $G$ is solvable if there exists some nonnegative integer $n$ for which $D^n(G)=\{e\}$, where $D^k(G)$ is the $k$th term of the derived series of $G$. The least integer $n$ satisfying this condition is called the solvability class of $G$. A group is abelian if and only if its solvability class is at most one; it is trivial if and only if its solvability class is zero.

Every nilpotent group is solvable. In particular, if a group is nilpotent of class at most $2^n-1$, then it is solvable of class at most $n$.

However, the converse is not true in general. For instance, $S_3$ is solvable of class 2: the first three terms of its derived series are $$S_3, \{ e, (123), (132)\}, \{e\} .$$ But it is not nilpotent: the terms of its lower central series are $$S_3, \{ e, (123), (132) \}, \{ e, (123), (132) \} , \dotsc .$$ In fact, $S_3$ is not even residually nilpotent, i.e., the infinite extension of the lower central series of $S_3$ never reduces to $\{e\}$.

In 1962, Walter Feit and John Thompson proved that every finite group of odd order is solvable (see Feit-Thompson Theorem). This result arose from a conjecture of William Burnside, and earlier work by Michio Suzuki.

## Characteristics of Solvable Groups

Proposition. Let $G$ be a group, and let $n$ be a positive integer. Then the following four conditions are equivalent.

1. $G$ is solvable of class at most $n$;
2. There exists a decreasing sequence $(G^k)_{0\le k \le n}$ of normal subgroups of $G$ such that $G = G_0$, $G^n=\{e\}$, and $G^k/G^{k+1}$ is abelian for every index $k;
3. There exists a decreasing sequence $(G^k)_{0\le k \le n}$ of subgroups of $G$ such that $G^0 = G$, $G^n = \{e\}$, $G^k$ normalizes $G^{k+1}$, and the quotient group $G^k/G^{k+1}$ is abelian for every index $k$;
4. There exists an abelian normal subgroup $N$ of $G$ for which $G/N$ is solvable of class at most $n-1$.

Proof. To show that (1) implies (2), we may take $G^k = D^k(G)$. Also, (2) evidently implies (3). To show that (3) implies (1), we note by induction that $D^k(G) \subseteq G^k$, for each index $k$; hence $D^n(G) \subseteq \{e\}$.

To show that (1) implies (4), we may take $N = D^{n-1}(G)$. To show that (4) implies (1), we define $\phi$ to be the canonical homomorphism from $G$ to $G/N$. Then $D^{n-1}(G) \subseteq \phi^{-1}(D^{n-1}(G/N)) = N$; since $N$ is commutative, $D^{n}(G) = (N,N) = \{e\}$. This completes the proof. $\blacksquare$

Thus a group is solvable if and only if it can be obtained by iterative extension by abelian groups.

Corollary. A finite group is solvable if and only if every quotient of its Jordan-Hölder series is a cyclic group of prime order.

Proof. A finite simple group is abelian if and only if it is cyclic and of prime order. Thus if the quotient groups of a Jordan-Hölder series of a group $G$ are cyclic and of prime order, then $G$ satisfies condition (3) of the proposition and hence is solvable.

Conversely, if $G$ is solvable, then it has a composition series whose quotients are abelian. Hence the quotients of the Jordan-Hölder series derived from this composition series are abelian, so they are cyclic and of prime order. $\blacksquare$

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