# For abelian groups: does knowing $\text{Hom}(X,Z)$ for all $Z$ suffice to determine $X$?

Let $X$ and $Y$ be abelian groups. Suppose $\text{Hom}(X,Z)\cong \text{Hom}(Y,Z)$ for all abelian groups $Z$. Does it follow that $X \cong Y$?

It has been answered before that this is true if the bijection $\text{Hom}(X,Z)\to \text{Hom}(Y,Z)$ is natural in $Z$. My intuition says that this assumption shouldn't be necessary. Maybe if we choose an extremely large and suitably "generic" group $Z$, then the structure of $\text{Hom}(X,Z)$ will somehow reveal the structure of $X$?

I'm also interested in the answer if "abelian group" is replaced by some other structure, in particular "$R$-module".

• My three cents: This is true for finitely generated abelian groups, questionable for general abelian groups, and almost certainly false for general $R$-modules. Feb 15 '15 at 1:23

I don't know how to answer this question for abelian groups, but I can answer it over some rings which have a very simple classification of all modules. In particular, suppose $$k$$ is a field and $$X$$ and $$Y$$ are $$k$$-vector spaces such that $$\operatorname{Hom}(X,Z)\cong\operatorname{Hom}(Y,Z)$$ (as $$k$$-vector spaces) for all $$k$$-vector spaces $$Z$$. Then it does follow that $$X\cong Y$$.
To prove this, suppose $$X\not\cong Y$$. Without loss of generality we may assume $$\dim X<\dim Y$$; let $$\kappa=\dim X$$ and $$\lambda=\dim Y$$. If $$\kappa$$ is finite the conclusion is trivial by taking $$Z=k$$, so suppose $$\kappa$$ is infinite. Let $$\mu$$ be a strong limit cardinal greater than or equal to $$|k|$$ and of cofinality $$\kappa^+$$ and let $$Z$$ be a $$k$$-vector space of dimension $$\mu$$. Then $$|Z|=\mu$$ so $$|\operatorname{Hom}(X,Z)|=\mu^\kappa$$ and $$|\operatorname{Hom}(Y,Z)|=\mu^\lambda$$. But $$\mu^\kappa=\mu$$ since $$\mu$$ is strong limit and $$\kappa<\operatorname{cf}(\mu)$$, whereas $$\mu^\lambda\geq\mu^{\kappa^+}=\mu^{\operatorname{cf}(\mu)}>\mu$$. Thus $$\operatorname{Hom}(X,Z)$$ and $$\operatorname{Hom}(Y,Z)$$ have different cardinalities, and so cannot be isomorphic.
We can generalize this a bit: the answer is also yes for $$R$$-modules if $$R$$ is a zero-dimensional principal ideal ring (in particular, for instance, this applies to $$R=\mathbb{Z}/(n)$$ for any $$n>0$$). Such a ring is a finite product of local zero-dimensional principal ideal rings, and so we may assume $$R$$ is local. In that case, $$R$$ has a nilpotent maximal ideal $$m$$, every ideal of $$R$$ is a power of $$m$$, and every $$R$$-module is a direct sum of cyclic modules.
So, any $$R$$-module has the form $$X\cong (R/m)^{\oplus\kappa_1}\oplus (R/m^2)^{\oplus \kappa_2}\oplus\dots\oplus (R/m^n)^{\oplus \kappa_n}$$ for cardinals $$\kappa_1,\dots,\kappa_n$$, where $$n$$ is minimal such that $$m^n=0$$. Now note that $$\operatorname{Hom}(R/m^i,R)\cong m^{n-i}R\cong R/m^i$$, so if $$Z=R^{\oplus\mu}$$ then $$\operatorname{Hom}(X,Z)\cong( (R/m)^{\oplus \mu})^{\kappa_1}\oplus ((R/m^2)^{\oplus \mu})^{\kappa_2}\oplus\dots\oplus ((R/m^n)^{\oplus\mu})^{\kappa_n}.$$ When $$\mu\geq|R|$$, $$((R/m^i)^{\oplus\mu})^\kappa$$ is isomorphic to $$(R/m^i)^{\oplus \mu^\kappa}$$ (proof sketch: an $$R$$-module $$N$$ is a direct sum of copies of $$R/m^i$$ iff $$\{x\in N:mx=0\}=m^{i-1}N$$, and this property is easy to verify for $$((R/m^i)^{\oplus\mu})^\kappa$$). Thus if you know $$\operatorname{Hom}(X,Z)$$ up to isomorphism for all $$Z$$, then you know the cardinals $$\mu^{\kappa_i}$$ for all $$\mu\geq|R|$$. As in the previous argument, this uniquely determines the cardinals $$\kappa_i$$, and so determines $$X$$ up to isomorphism.
• $|\operatorname{Hom}(X,Y)|=|k|×\dim Y^{\dim X}\neq\dim Y^{\dim X}$,where $k$ denotes the underlying field, so this answer need some modification. Jun 12 '20 at 1:35
• That doesn't matter because $\mu\geq|k|$. Jun 12 '20 at 1:45
• Surely that's not a big problem, just I think if you consider $\dim(\operatorname{Hom}(X,Z))$ instead of $|\operatorname{Hom}(X,Z)|$, the proof will become more elegant. Jun 12 '20 at 2:07