Inductive proof methods for the ascending series corresponding to a subgroup-defining function

Suppose $$f$$ is a subgroup-defining function and $$G$$ is a group. We can define an ascending series of subgroups of $$G$$ corresponding to $$f$$ as follows: $$M_0$$ is trivial, $$M_1 = f(G)$$, and each $$M_i$$ for $$i \ge 1$$ is such that $$M_i/M_{i-1} = f(G/M_{i-1})$$. In other words, the series is constructed by quotient-iteration. If we define $$q$$ as the quotient-defining function corresponding to $$f$$, then the quotient by $$M_i$$ can be identified with $$q^i(G) = q(q(\dots(q(G))\dots ))$$ (with the $$q$$ occurring $$i$$ times).

The purpose of this survey article is to consider the various inductive methods one might use to establish facts about the subgroups and quotients for the ascending series.

Using quotient-transitivity for subgroup properties
The following is specific enough to be used as a lemma in its own right, but obvious enough that the effort may not be justified:

If $$\alpha$$ is a subgroup property and $$f$$ is a subgroup-defining function satisfying the following conditions:


 * For any group $$G$$, $$f(G)$$ satisfies $$\alpha$$ in $$G$$
 * $$\alpha$$ is a quotient-transitive subgroup property: if $$H \le K \le G$$ are groups such that $$H$$ is a normal subgroup of $$G$$, $$H$$ satisfies $$\alpha$$ in $$G$$, and $$K/H$$ satisfies $$\alpha$$ in $$G/H$$, then $$K$$ satisfies $$\alpha$$ in $$G$$.

Then, for any group $$G$$, all the members of the ascending series for $$f$$ also satisfy $$\alpha$$.

Some examples are given below.

The proof can be executed inductively in any of a number of ways. In particular, it could be executed upward or downward. Explicitly, if the ascending series is $$(M_n)_{n \in \mathbb{N}_0}$$:


 * The upward method would move up the ascending series of a fixed group, using the inductive hypothesis and the fact of quotient-transitivity to go from one member of the ascending series to the next.
 * The downward method, if aiming to prove the result for $$n^{th}$$ member $$M_n$$, will induct on $$M_n/M_i$$ as a subgroup of $$G/M_i$$ with $$i$$ moving down from $$n$$ to $$0$$.

Using extension-closedness for group properties
The following is specific enough to be used as a lemma in its own right, but obvious enough that the effort may not be justified:

If $$\alpha$$ is a group property and $$f$$ is a subgroup-defining function satisfying the following conditions:


 * For any group $$G$$, $$f(G)$$ satisfies $$\alpha$$ as a group.
 * $$\alpha$$ is an extension-closed group property: if $$A$$ is a normal subgroup of a group $$B$$ such that both $$A$$ and $$B/A$$ satisfy $$\alpha$$ as groups, then $$B$$ satisfies $$\alpha$$.

Then, for any group $$G$$, all the members of the ascending series of $$G$$ corresponding to $$f$$ also satisfy $$\alpha$$.

The proof can be executed inductively in any of a number of ways. In particular, it could be executed upward or downward. Explicitly, if the ascending series is $$(M_n)_{n \in \mathbb{N}_0}$$::


 * The upward method would move up the ascending series of a fixed group, using the inductive hypothesis and the fact of quotient-transitivity to go from one member of the ascending series to the next.
 * The downward method, if aiming to prove the result for $$n^{th}$$ member $$M_n$$, will induct on $$M_n/M_i$$ as a subgroup of $$G/M_i$$ with $$i$$ moving down from $$n$$ to $$0$$.

Using extension-closedness or quotient-transitivity with respect to extensions of only a certain sort
In some cases, there are limitations on the sort of extension for which you can use the group property or subgroup property in the subgroup and quotient to deduce it for the extension.


 * The restriction may be of the sort that only allows for drawing conclusions in the case that the normal subgroup satisfies a property that $$f(G)$$ must satisfy. In the upper central series case, for instance, this might be the restriction to central extensions. In this case, the induction needs to be done downward, i.e., we will induct on $$M_n/M_i$$ as a subgroup of $$G/M_i$$ with $$i$$ moving down from $$n$$ to $$0$$.
 * The restriction may be of the sort that only allows for drawing conclusions in the case that the quotient group satisfies a property that $$f(G)$$ must satisfy. In this case, the induction needs to be done upward, i.e., we go from proving the result from $$M_i$$ to proving the result for $$M_{i+1}$$.