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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
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1 Answer

up vote 1 down vote accepted

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
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