Equivalence Between the Definitions of Solvable Group

Introduction

A group $G$ is solvable if there exists a chain of subgroups \(\lbrace e\rbrace =G_n\lhd G_{n-1}\lhd \cdots \lhd G_1 \lhd G_0 =G\) such that each $G_{i-1}/G_i$ is cyclic.

Theorem. For finite $G$, the definition of solvable is not altered when the condition that each quotient group is cyclic is replaced by being abelian.

Proof

One implication is immediate, since every cyclic group is abelian. Conversely, if each quotient is abelian, we will first focus on the case where $G$ is abelian. If it is, let $G_1$ be a maximal proper subgroup of $G$; then the abelian group $G/G_1$ has prime order. Otherwise, there would exist a proper nontrivial subgroup $H$ of $G/G_1$ such that its inverse image under the canonical projection from $G$ onto $G/G_1$ is a subgroup contained between $G_1$ and $G$, contradicting the maximality of $G_1$. Repeating the same argument on $G_1$, we obtain a subgroup $G_2$ of $G_1$ such that $G_1/G_2$ has prime order. Consequently, one obtains the desired chain of quotient groups with prime order, that is, cyclic¹. This process must eventually stop at the trivial subgroup $\lbrace e\rbrace $, since $G$ is finite.

Now, suppose that $G$ is a finite group with a normal subgroup $H$ such that $G/H$ is abelian. By the previous argument, there exists a chain of subgroups

\[\lbrace e\rbrace =K_n\lhd K_{n-1}\lhd \cdots \lhd K_1 \lhd K_0 =G/H\]

such that each $K_{i-1}/K_i$ is cyclic of prime order. Taking the canonical projection $\varphi:G\rightarrow G/H$ and defining $L_i=\varphi^{-1}(K_i)$, we obtain the chain of subgroups

\[H=L_n\lhd L_{n-1}\lhd\cdots\lhd L_1\lhd L_0=G.\]

Each $L_i$ is normal in $L_{i-1}$ because each $K_i$ is abelian and normal in its predecessors. Since $\varphi$ is surjective, $\varphi(L_i)=K_i$ for all $i$. Applying the Third Isomorphism Theorem to $\varphi$ restricted to $L_{i-1}$, we obtain that $L_{i-1}/L_i\cong \varphi(L_{i-1})/\varphi (L_i)$, that is,

\[L_{i-1}/L_i\cong K_{i-1}/ K_i,\]

from which it follows that $L_{i-1}/L_i$ is cyclic of prime order.

To conclude, if $G$ is a finite group with a chain of subgroups

\[\lbrace e\rbrace =N_n\lhd N_{n-1}\lhd \cdots \lhd N_1 \lhd N_0 =G\]

such that $N_{i-1}/N_i$ is abelian, as shown above, it is possible to refine this chain to obtain subchains between each $N_i$ and $N_{i-1}$ in such a way that the corresponding quotients are cyclic, and finally to construct a finer chain

\[\lbrace e\rbrace =G_n\lhd G_{n-1}\lhd \cdots \lhd G_1 \lhd G_0 =G\]

such that each $G_{i-1}/G_i$ is cyclic. $\blacksquare$

References

1. Lawrence, J. W., & Zorzitto, F. A. (2021). Abstract Algebra: A Comprehensive Introduction. Cambridge University Press. https://doi.org/10.1017/9781108874328