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I was thinking about the following: Say we got two groups $A$ and $B$, and we know that for any group C, there is a bijection

$$Hom(A,C) \to Hom(B,C).$$

Are $A$ and $B$ already isomorphic?

If the family of bijections was natural in $C$, this follows from Yoneda's lemma. But does it hold anyway without naturality, just from the cardinalities? I would suspect no but in simple cases (finitely generated). I'd like to hear any thoughts on how to come up with a proof or counterexample.

Also, does the answer change when we consider the contravariant case with an identification $Hom(C,A) \to Hom(C,B)$?

Dario
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2 Answers2

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Let $A$ be the free group with countably many generators, and let $B$ be group with countably many generators of which one has the relation $g^2=e$.

These are not isomorphic, because a free group is torsion-free and $B$ isn't.

If $C$ is any group with (finite or infinite) cardinality $\lambda$ of which $\kappa$ have order 1 or 2, then we have $$ \begin{align} |\operatorname{Hom}(A,C)| &= \lambda^{\aleph_0} \\ |\operatorname{Hom}(B,C)| &= \lambda^{\aleph_0}\cdot\kappa \end{align} $$ which are equal because $1\le \kappa \le \lambda $.

For the contravariant case, we can dualize Derek's idea: If there are injective homomorphisms $A\to B$ and $B\to A$, then $|\operatorname{Hom}(C,A)|=|\operatorname{Hom}(C,B)|$.

In that case we can take $A$ to be $\mathbb Q^\omega$ and $B=\mathbb Z\times\mathbb Q^\omega$. Then $A$ and $B$ are not isomorphic (because $B$ has an element that is not twice anything), but they easily inject into each other.

(As Derek notes in the comments you can also take $A$ and $B$ to be free groups of different finite ranks $\ge2$, for a finitely generated example. For example, if $A$ is the free group on $\{a,b\}$ and $B$ is the free group on $\{a,b,c\}$, then $A$ injects natively into $B$, and $\{a\mapsto ab, b\mapsto a^2b^2, c\mapsto a^3b^3\}$ generates an injective homomorphism $B\to A$).

darij grinberg
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hmakholm left over Monica
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  • Thanks, I thought that one would need some non-finitely generated group and your counterexample elaborates this perfectly. You might also post your answer under http://math.stackexchange.com/questions/1148289. – Dario May 31 '15 at 10:52
  • @Dario: I found a counterexample for the contravariant case too. – hmakholm left over Monica May 31 '15 at 11:32
  • Awesome! Thanks you two – Dario May 31 '15 at 12:28
  • So for a finitely geenrated example in the contravariant case you could take two free groups of different ranks. – Derek Holt May 31 '15 at 13:40
  • @DerekHolt: Can a free group inject into one with lower rank? It seems to me that would contradict the definition of rank. – hmakholm left over Monica May 31 '15 at 13:45
  • Yes, with free groups, the rank actually gets larger with increasing index! By the Schreier Index Formula, a subgroup of index $n$ in the free group of rank $r$ is free of rank $n(r-1)+1$. – Derek Holt May 31 '15 at 13:48
  • @DerekHolt: Sorry, you're right. I had it confused with rank of abelian groups. – hmakholm left over Monica May 31 '15 at 13:56
  • Of course, you don't need any cardinal arithmetics in the first part of the answer. It suffices to remark that $A$ and $B$ admit epimorphisms $A \to B$ and $B \to A$. Hence, we get monomorphisms $\hom(B,C) \to \hom(A,C)$ and $\hom(A,C) \to \hom(B,C)$. By Cantor-Schröder-Bernstein, it follows that $\hom(A,C) \cong \hom(B,C)$. – Martin Brandenburg Jun 03 '15 at 07:53
  • @MartinBrandenburg: Yes, that is the point of Derek Holt's answer. (And a perfectly good point, but I didn't want to plagiarize it here after he pointed it out). – hmakholm left over Monica Jun 03 '15 at 10:47
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If there is a surjective homomorphism $\rho:A \rightarrow B$ then, for any group $C$, $\alpha \mapsto \alpha \circ \rho$ is an injective map ${\rm Hom}(B,C) \to {\rm Hom}(A,C)$.

So, for any pair of groups $A$ and $B$ which are isomorphic to homomorphic images of each other, we have $|{\rm Hom}(A,C)| = |{\rm Hom}(B,C)|$.

There are examples of pairs of non-isomorphic finitely generated groups with this property. An example is given in Theorem 3 of the well-known paper "Some two-generator one-relator non-Hopfian groups" by Baumslag and Solitar

$$A = \langle a,b \mid a^{-1}b^2a=b^3 \rangle,$$ $$B = \langle c,d \mid c^{-1}d^2c=d^3, ([c,d]^2c^{-1})^2=1 \rangle.$$

darij grinberg
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Derek Holt
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