# Why $Hom(\mathbb{Q},G)=0$?

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.

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So $I_{2^n}$ means $\Bbb Z/2^n\Bbb Z$? Is $\langle a\rangle$ viewed as a subgroup of $F$, or of $G$? –  anon Dec 9 '12 at 8:12
$\langle a \rangle$ is a group in itself, isomorphic to $\mathbb{Z}$. –  Bombyx mori Dec 9 '12 at 8:44

## 1 Answer

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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Thanks. Let me sleep and think about the SES. –  Bombyx mori Dec 9 '12 at 9:16
I think there is a typo above. You mean $Hom(\mathbb{Q},\mathbb{Z})=0$ as a $\mathbb{Z}$-module homomorphism? I do not really get why the second homomorphisms is 0. –  Bombyx mori Dec 9 '12 at 10:36
I assume you're asking why $\text{Hom}(\mathbb{Q},\bigoplus \mathbb{Z}_{2^n})=0$? Take any homomorphism $f:\mathbb{Q}\to\bigoplus\mathbb{Z}_{2^n}$ and note that for any $x\in\mathbb{Q}$, $f(x)=2^nf(\frac{x}{2^n})$ and so you can show that any coordiante of $f(x)$ is zero. –  Alex Youcis Dec 9 '12 at 16:32
Can I ask again how to derive the exact sequence? I am still slightly lost. –  Bombyx mori Dec 23 '12 at 12:14