Let $G$ be the set of finitely generated groups up to isomorphism hence its elements will be noted $[B]$ where $B$ is some finitely generated group.

On this set we put a relation $\mathcal{ND}$ ("non-discernability") defined this way, let $[B]$ and $[C]$ be two elements in $G$ then :

$$[B]\mathcal{ND}[C]\Leftrightarrow \exists [A]\in G\text{ such that } [A\times B]=[A\times C]$$

One can show that the relation $\mathcal{ND}$ is a well defined equivalence relation. Few remarks have to be made about this relation, they come from those two posts on ME :



1 If $G$ is the set of groups up to isomorphism then the relation is trivial. That is two elements of $G$ are always non-discernable from each others.

2 On the other hand, if $G$ is the set of finite groups up to isomorphism then the relation is again trivial. The Krull-Schmidt theorem implies that for finite groups $A\times B$ isomorphic to $A\times C$ implies that $B$ is isomorphic to $C$. In this case the $\mathcal{ND}$ relation is just the equality on $G$.

3 In our case it is easily seen that if $[B]\mathcal{ND} [C]$ then $B^{ab}$ (the abelianization) is isomorphic to $C^{ab}$. Hence there are many different $\mathcal{ND}$-classes. Furthermore there exists a non-trivial finitely generated group $K$ such that $K\times K$ is isomorphic to $K$ hence we have $[K]\mathcal{ND}[\mathbb{I}]$ without having $[K]=[\mathbb{I}]$. Those two arguments show that the relation $\mathcal{ND}$ is not trivial.

Now come the different questions :

a. Can we find a $[B]\in G$ such that its $\mathcal{ND}$-class only have one element ? If yes, can we characterize them ?

Please note that the existence of $[K]\in G$ such that $[K^2]=[K]$ does imply that $[K\times B]\mathcal{ND}[B]$ hence $[K\times B]$ and $[B]$ are always in the same class but there is no reason, a priori, for them to be two different elements in $G$.

b. Can we compute the $\mathcal{ND}$-class of the trivial group $[\mathbb{I}]$ ?

Please note that any $[K]\in G$ such that $[K^2]=[K]$ is in the $\mathcal{ND}$-class of $[\mathbb{I}]$, but an element of the $\mathcal{ND}$-class of $[\mathbb{I}]$ has no reason, a priori, to be isomorphic to its square.

c. Being understood that the abelianization is a good invariant for $\mathcal{ND}$-classes are there more invariants?

Of course, I understand that this problem is very difficult and I will be happy even with a very partial answer or remark about this problem.


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