# Extending group homomorphisms

I was attempting to prove that for an exact sequence $$A_1\overset{\varphi_1}{\to}A_2\overset{\varphi_2}{\to}A_3...$$, the sequence $$...\hom(A_3,\mathbb{R})\overset{\varphi_2^*}{\to}\hom(A_2,\mathbb{R})\overset{\varphi_1^*}{\to}\hom(A_1,\mathbb{R})$$ is also exact, where $$\varphi_k^*(f)$$ is the homomorphism given by $$\varphi_k^*(f)(x)=f(\varphi_k(x))$$.

When showing that if $$\varphi_k^*(f)=0$$, then $$f=\varphi_{k+1}^*(g)$$, I simply let $$g$$ be the homomorphism defined on the image of $$\varphi_{k+1}$$ as $$g(x)=f(y)$$ if $$x=\varphi_{k+1}(y)$$. This is well defined because if $$z$$ is such that $$x=\varphi_{k+1}(z)$$, then $$\varphi_{k+1}(y-z)=0$$, so $$\varphi_k(w)=y-z$$ and $$f(y-z)=f(\varphi_k(w))=\varphi_k^*(f)(w)=0$$.

Now my problem is to extend $$g$$ to all of $$A_{k+2}$$, not just on the image $$\varphi_{k+1}$$. All groups are abelian.

My question is now, given an abelian group $$G$$ and a subgroup $$H$$ and a homomorphism $$f:H\to \mathbb{R}$$ is it always possible to extend $$f$$ to a homomorphism on all of $$G$$? I have managed to prove it for infinite cyclic groups $$G$$, but nothing in general, any hints?

Edit:

For finite cyclic groups it also holds since the only homomorphism from such a group to $$\mathbb{R}$$ is the trivial one.

• Maybe the transfer map $Ver:G/[G,G]\to H/[H,H]$ may help? Aug 22, 2021 at 10:28
• @DavidA.Craven Sorry, I meant infinite cyclic groups. Aug 22, 2021 at 13:11
• @DavidA.Craven I haven't checked the infinite cyclic group now, but how is your first map a homomorphism? $\phi(x^2x^2)=0$ but $\phi(x^2)+\phi(x^2)=-2$. Am I making a mistake? Aug 22, 2021 at 13:18
• Let $f:H\rightarrow\mathbb{R}$ and $g\notin H$. We can assume that $G=\langle g,H\rangle$. Then $G=\langle g\rangle+H$. (We consider $G$ to be an additive group). Then we consider two cases: 1) $\langle g\rangle\cap H=\{0\}$. In this case we take $f(g)=0$. 2) $kg\in H$ and $k$ a minimal positive integer with this property. Then each element $x\in G$ is uniquely represented as $x=lg+h$, where $0\leq l<k$ and $h\in H$. We can then put $f(g)=f(kg)/k$ and $f(x)=lf(g)+f(h)$. Wouldn't it work out that way. Aug 22, 2021 at 14:07
• @kabenyuk that is fair, but how can you assume $G=\langle g,H\rangle$? Aug 22, 2021 at 14:51