Solvable group

For example, the smallest Galois field extension of ${\displaystyle \mathbb {Q} }$ containing the element

${\displaystyle a={\sqrt[{5}]{{\sqrt {2}}+{\sqrt {3}}}}}$

gives a solvable group. It has associated field extensions

${\displaystyle \mathbb {Q} \subseteq \mathbb {Q} ({\sqrt {2}})\subseteq \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\subseteq \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\left(e^{2i\pi /5}\right)\subseteq \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\left(e^{2i\pi /5},a\right)}$

giving a solvable group of Galois extensions containing the following composition factors:

• ${\displaystyle \mathrm {Aut} \left(\mathbb {Q({\sqrt {2}})} \right/\mathbb {Q} )\cong \mathbb {Z} /2}$ with group action ${\displaystyle f\left(\pm {\sqrt {2}}\right)=\mp {\sqrt {2}},\ f^{2}=1}$, and minimal polynomial ${\displaystyle x^{2}-2}$.
• ${\displaystyle \mathrm {Aut} \left(\mathbb {Q({\sqrt {2}},{\sqrt {3}})} \right/\mathbb {Q({\sqrt {2}})} )\cong \mathbb {Z} /2}$ with group action ${\displaystyle g\left(\pm {\sqrt {3}}\right)=\mp {\sqrt {3}},\ g^{2}=1}$, and minimal polynomial ${\displaystyle x^{2}-3}$.
• ${\displaystyle \mathrm {Aut} \left(\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\left(e^{2i\pi /5}\right)/\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\right)\cong \mathbb {Z} /4}$ with group action ${\displaystyle h^{n}\left(e^{2im\pi /5}\right)=e^{2(n+1)mi\pi /5},\ 0\leq n\leq 3,\ h^{4}=1}$, and minimal polynomial ${\displaystyle x^{4}+x^{3}+x^{2}+x+1=(x^{5}-1)/(x-1)}$ containing the 5^th roots of unity excluding ${\displaystyle 1}$.
• ${\displaystyle \mathrm {Aut} \left(\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\left(e^{2i\pi /5},a\right)/\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})\left(e^{2i\pi /5}\right)\right)\cong \mathbb {Z} /5}$ with group action ${\displaystyle j^{l}(a)=e^{2li\pi /5}a,\ j^{5}=1}$, and minimal polynomial ${\displaystyle x^{5}-\left({\sqrt {2}}+{\sqrt {3}}\right)}$.

, where ${\displaystyle 1}$ is the identity permutation. All of the defining group actions change a single extension while keeping all of the other extensions fixed. For example, an element of this group is the group action ${\displaystyle fgh^{3}j^{4}}$. A general element in the group can be written as ${\displaystyle f^{a}g^{b}h^{n}j^{l},\ 0\leq a,b\leq 1,\ 0\leq n\leq 3,\ 0\leq l\leq 4}$ for a total of 80 elements.

It is worthwhile to note that this group is not abelian itself. For example:

${\displaystyle hj(a)=h(e^{2i\pi /5}a)=e^{4i\pi /5}a}$

${\displaystyle jh(a)=j(a)=e^{2i\pi /5}a}$

In fact, in this group, ${\displaystyle jh=hj^{3}}$. The solvable group is isometric to ${\displaystyle (\mathbb {C} _{5}\rtimes _{\varphi }\mathbb {C} _{4})\times (\mathbb {C} _{2}\times \mathbb {C} _{2}),\ \mathrm {where} \ \varphi _{h}(j)=hjh^{-1}=j^{2}}$, defined using the semidirect product and direct product of the cyclic groups. In the solvable group, ${\displaystyle \mathbb {C} _{4}}$ is not a normal subgroup.

The basic example of solvable groups are abelian groups. They are trivially solvable since a subnormal series is formed by just the group itself and the trivial group. But non-abelian groups may or may not be solvable.

More generally, all nilpotent groups are solvable. In particular, finite p-groups are solvable, as all finite p-groups are nilpotent.

In particular, the quaternion group is a solvable group given by the group extension

${\displaystyle 1\to \mathbb {Z} /2\to Q\to \mathbb {Z} /2\times \mathbb {Z} /2\to 1}$

where the kernel ${\displaystyle \mathbb {Z} /2}$ is the subgroup generated by ${\displaystyle -1}$.

Group extensions form the prototypical examples of solvable groups. That is, if ${\displaystyle G}$ and ${\displaystyle G'}$ are solvable groups, then any extension

${\displaystyle 1\to G\to G''\to G'\to 1}$

defines a solvable group ${\displaystyle G''}$. In fact, all solvable groups can be formed from such group extensions.

A small example of a solvable, non-nilpotent group is the symmetric group S₃. In fact, as the smallest simple non-abelian group is A₅, (the alternating group of degree 5) it follows that every group with order less than 60 is solvable.

The Feit–Thompson theorem states that every finite group of odd order is solvable. In particular this implies that if a finite group is simple, it is either a prime cyclic or of even order.

The group S₅ is not solvable — it has a composition series {E, A₅, S₅} (and the Jordan–Hölder theorem states that every other composition series is equivalent to that one), giving factor groups isomorphic to A₅ and C₂; and A₅ is not abelian. Generalizing this argument, coupled with the fact that A_n is a normal, maximal, non-abelian simple subgroup of S_n for n > 4, we see that S_n is not solvable for n > 4. This is a key step in the proof that for every n > 4 there are polynomials of degree n which are not solvable by radicals (Abel–Ruffini theorem). This property is also used in complexity theory in the proof of Barrington's theorem.

Consider the subgroups

${\displaystyle B=\left\{{\begin{bmatrix}*&*\\0&*\end{bmatrix}}\right\}{\text{, }}U=\left\{{\begin{bmatrix}1&*\\0&1\end{bmatrix}}\right\}}$ of ${\displaystyle GL_{2}(\mathbb {F} )}$

for some field ${\displaystyle \mathbb {F} }$. Then, the group quotient ${\displaystyle B/U}$ can be found by taking arbitrary elements in ${\displaystyle B,U}$, multiplying them together, and figuring out what structure this gives. So

${\displaystyle {\begin{bmatrix}a&b\\0&c\end{bmatrix}}\cdot {\begin{bmatrix}1&d\\0&1\end{bmatrix}}={\begin{bmatrix}a&ad+b\\0&c\end{bmatrix}}}$

Note the determinant condition on ${\displaystyle GL_{2}}$ implies ${\displaystyle ac\neq 0}$, hence ${\displaystyle \mathbb {F} ^{\times }\times \mathbb {F} ^{\times }\subset B}$ is a subgroup (which are the matrices where ${\displaystyle b=0}$). For fixed ${\displaystyle a,b}$, the linear equation ${\displaystyle ad+b=0}$ implies ${\displaystyle d=-b/a}$, which is an arbitrary element in ${\displaystyle \mathbb {F} }$ since ${\displaystyle b\in \mathbb {F} }$. Since we can take any matrix in ${\displaystyle B}$ and multiply it by the matrix

${\displaystyle {\begin{bmatrix}1&d\\0&1\end{bmatrix}}}$

with ${\displaystyle d=-b/a}$, we can get a diagonal matrix in ${\displaystyle B}$. This shows the quotient group ${\displaystyle B/U\cong \mathbb {F} ^{\times }\times \mathbb {F} ^{\times }}$.

Notice that this description gives the decomposition of ${\displaystyle B}$ as ${\displaystyle \mathbb {F} \rtimes (\mathbb {F} ^{\times }\times \mathbb {F} ^{\times })}$ where ${\displaystyle (a,c)}$ acts on ${\displaystyle b}$ by ${\displaystyle (a,c)(b)=ab}$. This implies ${\displaystyle (a,c)(b+b')=(a,c)(b)+(a,c)(b')=ab+ab'}$. Also, a matrix of the form

${\displaystyle {\begin{bmatrix}a&b\\0&c\end{bmatrix}}}$

corresponds to the element ${\displaystyle (b)\times (a,c)}$ in the group.

For a linear algebraic group ${\displaystyle G}$ its Borel subgroup is defined as a subgroup which is closed, connected, and solvable in ${\displaystyle G}$, and it is the maximal possible subgroup with these properties (note the second two are topological properties). For example, in ${\displaystyle GL_{n}}$ and ${\displaystyle SL_{n}}$ the group of upper-triangular, or lower-triangular matrices are two of the Borel subgroups. The example given above, the subgroup ${\displaystyle B}$ in ${\displaystyle GL_{2}}$ is the Borel subgroup.

In ${\displaystyle GL_{3}}$ there are the subgroups

${\displaystyle B=\left\{{\begin{bmatrix}*&*&*\\0&*&*\\0&0&*\end{bmatrix}}\right\},{\text{ }}U_{1}=\left\{{\begin{bmatrix}1&*&*\\0&1&*\\0&0&1\end{bmatrix}}\right\}}$

Notice ${\displaystyle B/U_{1}\cong \mathbb {F} ^{\times }\times \mathbb {F} ^{\times }\times \mathbb {F} ^{\times }}$, hence the Borel group has the form

${\displaystyle U\rtimes (\mathbb {F} ^{\times }\times \mathbb {F} ^{\times }\times \mathbb {F} ^{\times })}$

In the product group ${\displaystyle GL_{n}\times GL_{m}}$ the Borel subgroup can be represented by matrices of the form

${\displaystyle {\begin{bmatrix}T&0\\0&S\end{bmatrix}}}$

where ${\displaystyle T}$ is an ${\displaystyle n\times n}$ upper triangular matrix and ${\displaystyle S}$ is a ${\displaystyle m\times m}$ upper triangular matrix.

Any finite group whose p-Sylow subgroups are cyclic is a semidirect product of two cyclic groups, in particular solvable. Such groups are called Z-groups.

As a strengthening of solvability, a group G is called supersolvable (or supersoluble) if it has an invariant normal series whose factors are all cyclic. Since a normal series has finite length by definition, uncountable groups are not supersolvable. In fact, all supersolvable groups are finitely generated, and an abelian group is supersolvable if and only if it is finitely generated. The alternating group A₄ is an example of a finite solvable group that is not supersolvable.

If we restrict ourselves to finitely generated groups, we can consider the following arrangement of classes of groups:

cyclic < abelian < nilpotent < supersolvable < polycyclic < solvable < finitely generated group.

A group G is called virtually solvable if it has a solvable subgroup of finite index. This is similar to virtually abelian. Clearly all solvable groups are virtually solvable, since one can just choose the group itself, which has index 1.

A solvable group is one whose derived series reaches the trivial subgroup at a finite stage. For an infinite group, the finite derived series may not stabilize, but the transfinite derived series always stabilizes. A group whose transfinite derived series reaches the trivial group is called a hypoabelian group, and every solvable group is a hypoabelian group. The first ordinal α such that G^(α) = G^(α+1) is called the (transfinite) derived length of the group G, and it has been shown that every ordinal is the derived length of some group (Malcev 1949).