TITLE: Why $Hom(\mathbb{Q},G)=0$?
QUESTION [0 upvotes]: Let $F$ be the free abelian group generated by $X$, where $X=\{a,b_{1},b_{2},...,b_{n}...\}$ and let $K=\langle Y \rangle$. Here $Y=\{2a, a-2^{n}b_{n},n\ge 1\}$. Define $G=F/K$. Now I am supposed to prove:
$$0\rightarrow \langle a \rangle \rightarrow G\rightarrow \bigoplus_{n\ge 1}I_{2^{n}}\rightarrow 0$$
I am not sure how to construct this map. For example, with $2a\approx 0$ we should have the inclusion map to be $na\rightarrow n\pmod{2}a$. But this cannot be injective. 
A better choice is to map $a=2^{n}b_{n}$ to individual coordinates, since $2a=1$ we effectively map $a$ to $2^{n-1}$ in all $I_{2^{n}}$ coordinates. Now let the second map be the projection map that projects arbitrary element $g\in G$ to individual coordinates. This map is clearly surjective. But if I choose as above, then the sequence will not be exact. What is a good choice?
Then I am suppose to prove $$Hom(\mathbb{Q},G)=0$$ and I do not know how to prove it.

REPLY [1 votes]: You're almost there with the construction of the SES, keep working at it.
Once you have that there exists such an SES, the next observation is that since $\mathbb{Q}$ is divisible, it is an injective $\mathbb{Z}$-module. Thus, you see that applying $\text{Hom}(\mathbb{Q},\bullet)$ to
$$0\to\langle a\rangle\to G\to\bigoplus\mathbb{Z}_{2^n}\to 0$$
induces the SES
$$0\to\text{Hom}(G,\langle a\rangle)\to\text{Hom}(\mathbb{Q},G)\to\text{Hom}\left(\mathbb{Q},\bigoplus\mathbb{Z}_{2^n}\right)\to0$$
Now, I think you can quickly prove that both $\text{Hom}(G,\langle a\rangle)$ and $\text{Hom}\left(\mathbb{Q},\bigoplus\mathbb{Z}_{2^n}\right)$ are zero, which gives you the desired fact $\text{Hom}(\mathbb{Q},G)=0$.
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