There seems to be a standard trick in group theory which is to show that a group has a quotient group which "just not" has some property.

To make things clear:

• let $\mathcal{P}$ be a group property which implies finite presentation and is inherited by quotients
• let $G$ be a finitely generated group without $\mathcal{P}$
• then $G$ has a quotient $Q$ which is "just not $\mathcal{P}$", meaning $Q$ does not have $\mathcal{P}$ but all of its proper quotient do.

Question: when (and why) was this trick introduced?

I've seen it with $\mathcal{P}$=polycyclic used in Groves 1978 [Soluble groups with every proper quotient polycyclic]. The argument below is from a paper of Breuillard [On uniform exponential growth for solvable groups]. But there are certainly a few other possibilities ($\mathcal{P}$= finite and $\mathcal{P}$=minimax are the first that come to mind)

Proof of the trick: Let $G$ be such a group and let $\mathcal{N}$ be the set of all normal subgroups $N$ of $G$, such that $G/N$ does not satisfy $\mathcal{P}$. Suppose $N_1 \subset N_2 \subset ... \subset N_i \subset ...$ is an increasing chain of subgroups from $\mathcal{N}$. And let $N$ be the union of all $N_i$’s. Then $N$ is a normal subgroup of $G$. If $G/N$ had $\mathcal{P}$, it would have a finite presentation $\langle x_1 , ..., x_n \mid r_1 , ..., r_m \rangle$. The finitely many relations $r_i$’s would belong to one of the $N_i$’s, say $N_{i_0}$. Hence $G/N_{i_0}$ would appear as a quotient of $G/N$, hence has $\mathcal{P}$, contradicting the assumption that $G/N_{i_0}$ does not satisfy $\mathcal{P}$. So $G/N$ does not satisty $\mathcal{P}$. It follows that we can apply Zorn’s lemma and obtain a maximal element $M$ in $\mathcal{N}$. Then clearly $G/M$ does not satisfy $\mathcal{P}$, while any proper quotient of it has $\mathcal{P}$.

There seems to be a standard trick in group theory which is to show that a group has a quotient group which "just not" has some property.

To make things clear:

• let $\mathcal{P}$ be a group property which implies finite presentation and is inherited by quotients
• let $G$ be a finitely generated group without $\mathcal{P}$
• then $G$ has a quotient $Q$ which is "just not $\mathcal{P}$", meaning $Q$ does not have $\mathcal{P}$ but all of its proper quotient do.

Question: when (and why) was this trick introduced?

I've seen it with $\mathcal{P}$=polycyclic used in Groves 1978 [Soluble groups with every proper quotient polycyclic]. The argument below is from a paper of Breuillard [On uniform exponential growth for solvable groups]. But there are certainly a few other possibilities ($\mathcal{P}$= finite is the first that comes to mind)

Proof of the trick: Let $G$ be such a group and let $\mathcal{N}$ be the set of all normal subgroups $N$ of $G$, such that $G/N$ does not satisfy $\mathcal{P}$. Suppose $N_1 \subset N_2 \subset ... \subset N_i \subset ...$ is an increasing chain of subgroups from $\mathcal{N}$. And let $N$ be the union of all $N_i$’s. Then $N$ is a normal subgroup of $G$. If $G/N$ had $\mathcal{P}$, it would have a finite presentation $\langle x_1 , ..., x_n \mid r_1 , ..., r_m \rangle$. The finitely many relations $r_i$’s would belong to one of the $N_i$’s, say $N_{i_0}$. Hence $G/N_{i_0}$ would appear as a quotient of $G/N$, hence has $\mathcal{P}$, contradicting the assumption that $G/N_{i_0}$ does not satisfy $\mathcal{P}$. So $G/N$ does not satisty $\mathcal{P}$. It follows that we can apply Zorn’s lemma and obtain a maximal element $M$ in $\mathcal{N}$. Then clearly $G/M$ does not satisfy $\mathcal{P}$, while any proper quotient of it has $\mathcal{P}$.

There seems to be a standard trick in group theory which is to consider show that a group has a quotient group which is "just not" have some property.

To make things clear:

• let $\mathcal{P}$ be a group property which implies finite presentation
• let $G$ be a finitely generated group which has a quotient without $\mathcal{P}$
• then $G$ has a quotient $Q$ which is "just not $\mathcal{P}$", meaning $Q$ does not have $\mathcal{P}$ but all of its proper quotient do.

Question: when (and why) was this trick introduced?

I've seen it with $\mathcal{P}$=polycyclic used in Groves 1978 [Soluble groups with every proper quotient polycyclic]. The argument below is from a paper of Breuillard [On uniform exponential growth for solvable groups]. But there are certainly a few other possibilities ($\mathcal{P}$= finite and $\mathcal{P}$=hyperbolic are the first that come to mind)

Proof of the trick: Let $G$ be such a group and let $\mathcal{N}$ be the set of all normal subgroups $N$ of $G$, such that $G/N$ does not satisfy $\mathcal{P}$. Suppose $N_1 \subset N_2 \subset ... \subset N_i \subset ...$ is an increasing chain of subgroups from $\mathcal{N}$. And let $N$ be the union of all $N_i$’s. Then $N$ is a normal subgroup of $G$. If $G/N$ had $\mathcal{P}$, it would have a finite presentation $\langle x_1 , ..., x_n \mid r_1 , ..., r_m \rangle$. The finitely many relations $r_i$’s would belong to one of the $N_i$’s, say $N_{i_0}$. Hence $G/N_{i_0}$ would appear as a quotient of $G/N$, hence has $\mathcal{P}$, contradicting the assumption that $G/N_{i_0}$ does not satisfy $\mathcal{P}$. So $G/N$ does not satisty $\mathcal{P}$. It follows that we can apply Zorn’s lemma and obtain a maximal element $M$ in $\mathcal{N}$. Then clearly $G/M$ does not satisfy $\mathcal{P}$, while any proper quotient of it has $\mathcal{P}$.

There seems to be a standard trick in group theory which is to show that a group has a quotient group which "just not" has some property.

To make things clear:

• let $\mathcal{P}$ be a group property which implies finite presentation and is inherited by quotients
• let $G$ be a finitely generated group without $\mathcal{P}$
• then $G$ has a quotient $Q$ which is "just not $\mathcal{P}$", meaning $Q$ does not have $\mathcal{P}$ but all of its proper quotient do.

Question: when (and why) was this trick introduced?

I've seen it with $\mathcal{P}$=polycyclic used in Groves 1978 [Soluble groups with every proper quotient polycyclic]. The argument below is from a paper of Breuillard [On uniform exponential growth for solvable groups]. But there are certainly a few other possibilities ($\mathcal{P}$= finite and $\mathcal{P}$=minimax are the first that come to mind)

Proof of the trick: Let $G$ be such a group and let $\mathcal{N}$ be the set of all normal subgroups $N$ of $G$, such that $G/N$ does not satisfy $\mathcal{P}$. Suppose $N_1 \subset N_2 \subset ... \subset N_i \subset ...$ is an increasing chain of subgroups from $\mathcal{N}$. And let $N$ be the union of all $N_i$’s. Then $N$ is a normal subgroup of $G$. If $G/N$ had $\mathcal{P}$, it would have a finite presentation $\langle x_1 , ..., x_n \mid r_1 , ..., r_m \rangle$. The finitely many relations $r_i$’s would belong to one of the $N_i$’s, say $N_{i_0}$. Hence $G/N_{i_0}$ would appear as a quotient of $G/N$, hence has $\mathcal{P}$, contradicting the assumption that $G/N_{i_0}$ does not satisfy $\mathcal{P}$. So $G/N$ does not satisty $\mathcal{P}$. It follows that we can apply Zorn’s lemma and obtain a maximal element $M$ in $\mathcal{N}$. Then clearly $G/M$ does not satisfy $\mathcal{P}$, while any proper quotient of it has $\mathcal{P}$.