Exercise 3.4.8 (Characterization of finite solvable groups)

Proof. (of Ex. 3.4.8.) Let $G$ be a finite group.

(i) $⟺$(ii):

Suppose that $G$ has a chain of subgroups: $1=H_0⊴H_1⊴H_2\cdots ⊴H_s=G$ such that $H_{i+1}∕H_i$ is cyclic for $0\le i\le s-1$. Then $H_{i+1}∕H_i$ is abelian, so $G$ is solvable.

Conversely, suppose that $G$ is solvable, so that there is a chain of subgroups:

such that $G_{i+1}∕G_i$ is abelian.

By the lemma, applied to the abelian group $G_{i+1}∕G_i$ there are for every $i\in [[1,s]]$ some chain of subgroups ${({\bar{G}}_{i,k})}_{0\le k\le s_i}$:

such that ${\bar{G}}_{i,k+1}∕{\bar{G}}_{i,k}$ is cyclic of prime order.

By the Lattice Isomorphism Theorem, there are subgroups $G_{i,k}$ of $G$ such that ${\bar{G}}_{i,k}=G_{i,k}∕G_i$. By part (5) of this Theorem $G_{i,k}⊴G_{i,k+1}(0\le k<s_i)$, and by the Third Isomorphism Theorem, ${\bar{G}}_{i,k+1}∕{\bar{G}}_{i,k}=(G_{i,k+1}∕H)∕(G_{i,k}∕H)\simeq G_{i,k+1}∕G_{i,k}$, so $G_{i,k+1}∕G_{i,k}$ is simple of prime order. We obtain for every $i\in [[0,t[[$ the chain

whose quotients are cyclic of prime orders.

Gluing together these $t$ chains, we obtain a chain

whose quotients are cyclic of prime orders. The induction is done, which proves (ii).

(i) $⟺$(iii): If all composition factors are of prime orders, then these factors are cyclic. By the preceding equivalence, $G$ is solvable.

Conversely, if $G$ is solvable, we have proved above that $G$ has a chain of subgroups $1=H_0⊴H_1⊴H_2\cdots ⊴H_s=G$ such that $H_{i+1}∕H_i$ is cyclic of prime order, so $H_{i+1}∕H_i$ is simple. Then this chain is a Jordan-Hölder chain, whose composition factors are cyclic of prime order.

(i) $⟺$(iv): If (iv) is true, then $G$ is solvable by definition.

Conversely, suppose that a group $G$ is solvable. If $G$ is trivial, the chain has a unique element $\{1\}$. We suppose now that $G$ is not trivial. Following the hint, consider a minimal nontrivial normal subgroup $M$ (such a subgroup exists, because $G⊴G$, and $G$ is not trivial, then the set $S$ of nontrivial normal subgroup of $G$ is not vacuous, and finite, so $S$ has a minimal element).

By Exercise 5, $M$ is solvable, so by (iii) the penultimate element $N$ of the composition series of $M$ satisfies $N⊴M$ and $M∕N$ is cyclic of prime order $p$. A fortiori $M∕N$ is abelian, so

for all $x,y\in M$.

This gives $\mathit{𝑥𝑦𝑁}=\mathit{𝑦𝑥𝑁}$ and $x^{-1}{y}^{-1}\mathit{𝑥𝑦𝑁}=N$, so

Let $g$ be any element of $G$. Then ${\gamma }_g:G\to G$ defined by ${\gamma }_g(x)=\mathit{𝑔𝑥}{g}^{-1}$ is an automorphism of $G$. Since $N⊴M$, then ${\gamma }_g(N)⊴{\gamma }_g(M)=M$, so

Moreover $|\mathit{𝑔𝑁}{g}^{-1}|=|N|$, thus $|M∕\mathit{𝑔𝑁}{g}^{-1}|=|M∕N|=p$, so $\mathit{𝑔𝑁}{g}^{-1}$ is a normal subgroup of $M$ of index $p$ in $M$. The same argument as above shows that $x^{-1}{y}^{-1}\mathit{𝑥𝑦}\in \mathit{𝑔𝑁}{g}^{-1}$:

Therefore if $x,y\in M$,

Put $K=\underset{g\in G}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}\mathit{𝑔𝑁}{g}^{-1}$. Then $K⊴G$, and $K\subseteq N⊊M$. The minimality of $M$ among the nontrivial normal subgroups of $G$ shows that $K=\{1\}$, so

Hence $x^{-1}{y}^{-1}\mathit{𝑥𝑦}=1$, so $\mathit{𝑥𝑦}=\mathit{𝑦𝑥}$ for all elements $x,y\in M$: $M$ is abelian.

We conclude by complete induction on the order of $G$. The result is vacuously true for the trivial group. Assume that all groups of order smaller than $n$ satisfy (iv), and let $G$ be a group of order $n>1$. We know that there exists a non trivial abelian normal subgroup $M$ of $G$. If $G=M$, then the short chain $\{1\}⊴M=G$ proves (iv). Otherwise $|M|<n$, and the induction hypothesis shows that $G∕M$ satisfies (iv). There exists a chain:

where ${\bar{K}}_i⊴G∕M$ and ${\bar{K}}_{i+1}∕{\bar{K}}_i$ is abelian.

The Lattice Isomorphism Theorem shows the existence of subgroups $K_i$ of $G$ such that $M\le K_i$ and ${\bar{K}}_i=K_i∕M$. Then $K_{i+1}∕K_i\simeq {\bar{K}}_{i+1}∕\overline{K_i}$ is abelian ( $0\le i<t$), and $\overline{K_i}⊴G∕M$ implies $K_i⊴G$ ( $0\le i\le t$).

Since $M$ is abelian (and $K_0=M,K_t=G$), the chain

satisfies (iv). The induction is done which proves (iv) for all finite solvable groups $G$. □