# Show $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ for finite $A$

I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group. Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.

The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.

Some thoughts:

• I've tried to assume the existence of a nontrivial homorphism $A \to \mathbb{Q}$ to generate a contradiction but with no success.
• We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:

$$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$ Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all.

• I don't see much reason to involve the tensor product here. A non-zero map will give you an element of finite order in the rationals. – Tobias Kildetoft Nov 17 '16 at 11:08

## 2 Answers

I don't see how to prove this using tensor products the way you are trying to, but there is a very easy direct argument. Namely, notice that for every $a\in A$ there is some nonzero $n\in\mathbb{Z}$ such that $na=0$ (for instance, $n=|A|$). If $f:A\to\mathbb{Q}$ is a homomorphism, then $nf(a)=f(na)=0$ as well. Since $\mathbb{Q}$ is torsion-free, this means $f(a)=0$, and so $f=0$ since $a$ was arbitrary.

Note in particular that it is not true in general that $M\otimes N=0$ implies $\operatorname{Hom}(M,N)=0$. For instance, if $M=N=\mathbb{Q}/\mathbb{Z}$, we have $\mathbb{Q}/\mathbb{Z}\otimes\mathbb{Q}/\mathbb{Z}=0$, but $\operatorname{Hom}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})$ is nontrivial (for instance, it contains the identity map). So you will need to use something special about finite groups and/or $\mathbb{Q}$, instead of just using formal properties of tensors and Homs.

• I think the intended solution was to essentially use that the dual of a finite group is itself finite and then move over the group to the other side of the Hom. I am just not sure which dual to use for this. – Tobias Kildetoft Nov 17 '16 at 11:25

You can easily see that all spaces (except from $A$) involed in $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$

are $\Bbb Q$-vectorspaces. Than you only need to show that all of the group homomorphisms between the vector spaces are vector space homomorphisms. Thus you get $$0 = \text{Hom}_{\Bbb Q \textbf{-Vect}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) \cong \text{Hom}_{Ab}(A,\mathbb{Q}).$$

• @EricWofsey Did I fixed it now? – user60589 Nov 17 '16 at 11:38
• Yeah, that works and is a nice argument (though I doubt it was the argument intended by whoever posed the problem!). – Eric Wofsey Nov 17 '16 at 11:40