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When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old question but shaped in a different manner:

Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient $(A\ast B)/ncl(w^k)$ nontrivial, where $ncl(w^k)$ is the normal closure of $w^k$?

This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.

It is not true in general when $k =1$. For example, let $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$ and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particularly interested in the following question

When $k=1$, are there counterexamples if $A,B$ are torsion-free?

$\DeclareMathOperator\ncl{ncl}$When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old question but shaped in a different manner:

Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient $(A\ast B)/\ncl(w^k)$ nontrivial, where $\ncl(w^k)$ is the normal closure of $w^k$?

This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.

It is not true in general when $k =1$. For example, let $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$ and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particularly interested in the following question

When $k=1$, are there counterexamples if $A,B$ are torsion-free?

When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old question but shaped in a different manner:

Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient $(A\ast B)/ncl(w^k)$ nontrivial, where $ncl(w^k)$ is the normal closure of $w^k$?

This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.

It is not true in general when $k =1$. For example, let $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$ and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particular interested in the following question

When $k=1$, are there counterexamples if $A,B$ are torsion-free?

When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old question but shaped in a different manner:

Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient $(A\ast B)/ncl(w^k)$ nontrivial, where $ncl(w^k)$ is the normal closure of $w^k$?

This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.

It is not true in general when $k =1$. For example, let $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$ and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particularly interested in the following question

When $k=1$, are there counterexamples if $A,B$ are torsion-free?