# Abstract Nonsense

## Review of Group Theory: Solvable Groups

Point of post: In this post we introduce the notion of solvable groups and derive some simple facts about them.

Motivation

Solvable groups are things that at first may seem useless. They have a somewhat complicated definition that seems almost completely arbitrary–in fact the opposite is true. The notion of solvable groups arose in the study of the solvability by radicals of certain polynomial equations, in particular in Galois theory. Specifically it was discovered that to every polynomial is associated a group (called the Galois group) and moreover it was discovered that the polynomial was solvable if and only if there existed a point after which a certain ‘normal chain of subgroups’ terminated (in the sense that after that point all the groups in the chain are trivial). With such an utterly fascinating result it is clear that the precise definition of this aforementioned condition, sufficient conditions for groups to have this condition, and properties that such groups have is absolutely important. This notion is the notion of solvability.

Solvable Groups

Let $G$ be a group. Recall the definition of the commutator subgroup of a group and recursively define $G^{(n)}$ by $G^{(0)}=G$ and $G^{(n)}=\left[G^{(n-1)},G^{(n-1)}\right]$. We call then the series

$G^{(0)}\trianglerighteq G^{(1)}\trianglerighteq G^{(2)}\trianglerighteq\cdots$

the derived series of $G$ (noting that the normality of each subgroup having already been proven). We say that the derived series terminates if there exists some $N\in\mathbb{N}$ for which $n\geqslant N$ implies $G^{(n)}=\{e\}$. With this in mind we define a group $G$ to be solvable if its derived series terminates.

An important thing to note is that a non-trivial non-abelian simple group $G$ can never be solvable. Indeed, suppose that $G$ is non-abelian and simple. Note then that $G^{(1)}=\left[G,G\right]\unlhd G$ and so by assumption of simplicity one must conclude that $\left[G,G\right]=\{e\}$ or $\left[G,G\right]=G$. Now, since $G$ is non-abelian we know that $\left[G,G\right]\ne\{e\}$ and so $\left[G,G\right]=G$. It easily follows then by induction that $G^{(n)}=G\ne\{e\}$ for every $n\in\mathbb{N}$ and so $G$ is not simple. We summarize this in the slightly more (for us) interesting form:

Theorem: Let $G$ be a non-trivial non-abelian solvable group. Then, $G$ is not simple.

Our next result will give us an equivalent formulation of solvability. Namely:

Theorem: Let $G$ be a group. Then $G$ is solvable if and only if there exists a finite set $\left\{H^{(0)},\cdots,H^{(n)}\right\}$ of subgroups of $G$ such that

$\{e\}=H^{(n)}\unlhd H^{(n-1)}\unlhd\cdots\unlhd H^{(0)}=G$

and $H^{(i-1)}/H^{(i)}$ is abelian for each $i=1,\cdots,n$.

Proof: Assume first that $G$ is solvable. We’ve already seen in past posts that for any group $K$ one has that $\left[K,K\right]\unlhd G$ and so we get that $G^{(i)}\unlhd G^{(i-1)}$ for $i=1,\cdots,n$. Moreover, since $G^{(i-1)}/G^{(i)}$ is just the abelinization of $G^{(i-1)}$ it is evidently abelian. Since $G^{(n)}=\{e\}$ (if $G^{(n)}$ terminates at $n$) one has that $H^{(i)}=G^{(n)}$ satisfies the conditions and so the conclusion follows.

Conversely, suppose that such a series of $H^{(i)}$‘s existed. Then, we know by our previous knowledge of commutator subgroups that since $H^{(i-1)}/H^{(i)}$ is normal that $\left[H^{(i-1)},H^{(i-1)}\right]\subseteq H^{(i)}$. Thus, by induction we have that $G=G^{(0)}\subseteq H^{(0)}=G$ and so assume that $G^{(i)}\subseteq H^{(i)}$ then $G^{(i+1)}=\left[G^{(i)},G^{(i)}\right]\subseteq\left[H^{(i)},H^{(i)}\right]\subseteq H^{(i+1)}$ and the induction is complete. In particular we see that this implies that $G^{(n)}\subseteq H^{(n)}=\{e\}$ so that $G^{(n)}=\{e\}$ and thus the derived series of $G$ terminates. The conclusion follows. $\blacksquare$

Our first results give us some sufficiency conditions for when a group $G$ is solvable. This will enable us to conclude that a large array of groups (most notably $p$-groups) are solvable. In particular:

Theorem: A subgroup of a solvable group is solvable. Moreover, let $G$ be a group and $N\unlhd G$. Then, $G$ is solvable if and only if $N$ and $G/N$ are solvable.

Proof: Suppose first that $G$ is solvable and $H\leqslant G$. Let $\{e\}=H^{(n)}\unlhd\cdots\unlhd H^{(0)}=G$ be the sequence of subgroups such that $H^{(i-1)}/H^{(i)}$ is abelian. Define $K^{(i)}=H\cap H^{(i)}$. By first principles we know that $\{e\}=K^{(n)}\unlhd\cdots\unlhd K^{(0)}=H$. Moreover, we claim that $K^{(i-1)}/K^{(i)}$ is abelian for $i=1,\cdots,n$. Indeed, by the second isomorphism theorem we have that

$K^{(i-1)}/K^{(i)}=K^{(i-1)}/\left(H\cap H^{(i)}\right)\cong K^{(i-1)}H^{(i)}/H^{(i)}$

That said $K^{(i-1)}H^{(i)}=\left(H^{(i-1)}\cap H\right)H^{(i)}\subseteq H^{(i-1)}$ and so $K^{(i-1)}/K^{(i)}$ is isomorphic to a subgroup of an abelian group, and thus is clearly abelian. The conclusion follows.

Secondly, suppose that $N\unlhd G$ and $N$ and $G/N$ are both solvable. Note by the fourth isomorphism theorem if $\{N\}=Q^{(n)}\unlhd\cdots\unlhd Q^{(0)}=G/N$ there exists subgroups $N=P^{(n)}\unlhd\cdots\unlhd P^{(0)}=G$ moreover

$P^{(i-1)}/P^{(i)}\cong \left(P^{(i-1)}/N\right)/\left(P^{(i)}/N\right)=Q^{(i-1)}/Q^{(i)}$

and thus $P^{(i-1)}/P^{(i)}$ is abelian. But, since $N$ is solvable there exists $\{e\}=S^{(m)}\unlhd\cdots\unlhd S^{(0)}=N$. Thus, combining these two we get

$\{e\}=S^{(m)}\unlhd\cdots\unlhd S^{(0)}=N=P^{(n)}\unlhd\cdots\unlhd P^{(0)}=G$

from where the conclusion follows.

Conversely, by the first part of the theorem we know that $N$ is abelian so it remains to show that $G/N$ is solvable. More generally any homomorphic image of $G$ is solvable. To see this suppose that $\phi:G\twoheadrightarrow H$ is an epimorphism. Note then the trivial fact that if $\{e\}=G^{(n)}\unlhd\cdots\unlhd G^{(0)}=G$ is the derived series for $G$ that $\phi\left(G^{(i)}\right)\supseteq H^{(i)}$ where $H=H^{(0)}\trianglerighteq \cdots$ is the derived series for $H$ In particular $H^{(n)}\subseteq \phi\left(G^{(n)}\right)=\phi\left(\{e\}\right)=\{e\}$ and so $H^{(n)}=\{e\}$ and thus $H$ is solvable.$\blacksquare$

With this we can make a “basic” (in a sense made apparent by a later theorem) sufficiency condition for groups to solvable. Namely:

Theorem: Every $p$-group is solvable.

Proof: We prove this by induction. Namely fix an arbitrary prime $p$ and we induct on groups of order $p^m$. If $m=1$ then every group is abelian and thus trivially solvable. So, assume that every group of order $p,\cdots,p^m$ is prime and let $G$ be a group of order $p^{m+1}$. We know from first principles that $\mathcal{Z}(G)$ is non-trivial. If $G=\mathcal{Z}(G)=G$ and we’re done so assume not. Then, we have that $\mathcal{Z}(G)\unlhd G$, $\left|\mathcal{Z}(G)\right|=p^s\;\; s\leqslant m$, and $\left|G/\mathcal{Z}(G)\right|=p^r\;\; r\leqslant m$ and thus by the induction hypothesis $\mathcal{Z}(G)$ and $G/\mathcal{Z}(G)$ are solvable. But, by our previous theorem this implies that $G$ itself is solvable. Thus, the induction is complete. $\blacksquare$

References:

1. Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.

2. Hungerford, Thomas W. Algebra. Hungerford. New York: Rinehart and Winston, 1974. Print.

March 11, 2011 -