Let $G$ be the set of finitely generated groups up to isomorphism.

Now define $B$ and $C$ two finitely generated groups to be not discernable if one can find a finitely generated group $A$ such that $A\times B$ is isomorphic to $A\times C$. This defines an equivalence relation on $G$ (straightforward verification).

The questions I will ask are related to this post :

For groups $A,B,C$, if $A\times B$ and $A\times C$ are isomorphic do we have $B$ isomorphic to $C$?

In this post, from the counter-example for $(1)$ not every class is reduced to one (you can not always discern a group from another). From the counter-example to $(2)$ there are many classes (using the abelianization one can discern groups using edit 2). So the "non-discernability" is clearly a non-trivial relation (so this is different from both the finite groups case and the general groups case). I recall that we are in $G$ and hence all groups are finitely generated.

(3) Are there groups $B$ such that the "non-discernability" class is reduced to $B$ ? If yes, can we caracterize them ?

(4) Is the trivial group "non-discernability" class reduced to the trivial group? If not, can we caracterize groups within this class ?

($\infty$) Give invariants for the "non-discernability" classes.

  • 1
    $\begingroup$ I think your question would be more readable if you removed every reference to "isomorphism class". In general, this is assumed so you don't need to point it out (but perhaps keep your first line, so as to satisfy pedants!). $\endgroup$ – user1729 May 1 '15 at 9:25

I have a favourite paper. However, I have never had a chance to apply it until now!

The paper is Some odd finitely presented groups by Gilbert Baumslag and Charles F. Miller III. (Mostly I like the paper because of the title.) The paper, among other interesting things, constructs a finitely generated group $K$ such that $K\cong K\times K$ (Theorem D).

Therefore, $K\times 1\cong K\times K$, and more generally $K\times B\cong K\times (K\times B)$. Hence, there are no groups $B$ whose "non-discernability" class reduces to $B$. Not even the trivial group.

This answers (3), and part of (4). Groups in the "non-discernability" class of the trivial group are all finitely generated groups $Q$ with $Q\cong Q\times S$, $S$ some finitely generated group. Perhaps a good starting point for your reading would be the paper J. M. Tyrer Jones Direct products and the Hopf property, J. Austral. Math Soc. (1974), where they constructed the first finitely generated groups $Q$ such that $Q\cong Q\times Q$.

See also J. M. Tyrer Jones Direct products and the Hopf property, J. Austral. Math Soc. (1974) and David Meier, Non-Hopfian groups J. London Math. Soc., (1982)

  • $\begingroup$ Thanks for your answer, the paper is very interesting and you clearly killed the question (3). For the (4) question, all groups isomorphic to their squares are certainly in the trivial group's "non-discernability"-class. However it might happen that $A\times K$ is isomorphic to $A$ for some $A$ without having $K\times K$ isomorphic to $K$ . So the $\mathcal{ND}$-class of the trivial group might be bigger than the set of groups isomorphic to their squares, right? $\endgroup$ – Clément Guérin May 1 '15 at 10:07
  • $\begingroup$ Ah, sorry, of course. My mistake. I'll edit that in. $\endgroup$ – user1729 May 1 '15 at 10:12

It is an open problem whether there exists a finitely presented group $G$ satisfying $G \simeq G \times G$. However, there exist such finitely generated groups. I think the first example was given by Jones in Direct products and the Hopf property.


Since the previous answers provide negative result, I restrict to positive results. First, to be in keeping with standard terminology (very common in module theory, fibre bundles, etc), what you call "non-discernable" should rather be called "stably isomorphic" (within finitely generated groups).

An easy invariant of stable isomorphism: the (isomorphism type) of the abelianization. It immediately follows from the fact that $A\times B\simeq A\times C$ implies $B\simeq C$ if $A,B,C$ are finitely generated abelian groups.

A stronger result is the following:

if $B,C$ are stably isomorphic, then they have isomorphic profinite completions (that is, their profinite completions are isomorphic as topological groups).

Indeed, if $G$ is a group, let $G_n$ be the quotient of $G$ by the intersection of all subgroups of index $\le n$. If $G$ is finitely generated, then $G_n$ is finite, and the profinite completion of $G$ is naturally isomorphic (as a topological group) to the inverse limit of the $G_n$. A nice feature of this quotient is that $(A\times B)_n=A_n\times B_n$ for all groups $A,B$ (it is maybe easier to see if we define $K_n(G)$ to be the intersection of all subgroups of index $\le n$, then it is immediate that $K_n(A\times B)=K_n(A)\times K_n(B)$ for any groups $A,B$). Therefore if $A,B,C$ are finitely generated groups and $A\times B\simeq A\times C$, then for all $n$ we have $A_n\times B_n\simeq A_n\times C_n$, and then (Wedderburn-Remak-Krull-Schmidt) it follows that $B_n\simeq C_n$ for all $n$. This implies that $B$ and $C$ have the same finite quotients and a result of Ribes and Zaleskii (see https://mathoverflow.net/questions/39973/distinguishing-pro-finite-completions) implies that $B$ and $C$ have isomorphic profinite completions.

In particular if two finite groups are stably isomorphic then they are isomorphic.

It remains to think about other invariants and find two finitely generated groups with isomorphic profinite completions, but not stably isomorphic!


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