$0 \rightarrow \mathbb{Z} \stackrel{f}{\rightarrow} \mathbb{Z}^3 \stackrel{h}{\rightarrow} H_1(X) \rightarrow 0$
Where $f(1) = (1,0,2)$
Then $H_1(X) \cong \frac{\mathbb{Z}^3}{im(f)} \cong \frac{\mathbb{Z}^3}{\langle 1,0,2 \rangle}$
Is this all correct so far? The image of $1$ under $f$ is $(1,0,2)$, but i'm just confused. The correct answer is supposed to be $H_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}_2$.. I'm trying to think about the best way to make sense of this. I often have problems on steps like this when analyzing a short exact sequence. Maybe I should think of $im(g) = \mathbb{Z} \oplus 2\mathbb{Z}$ and then $\frac{\mathbb{Z}^3}{\mathbb{Z} \oplus 2\mathbb{Z}} \cong \frac{\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}}{\mathbb{Z} \oplus 2\mathbb{Z}} \cong \mathbb{Z} \oplus \mathbb{Z}_2$
Any insight in general is appreciated!! Thanks!
The group is isomorphic to $\mathbb{Z} ^2$. Define the homomorphism $f:\mathbb{Z}^3 \rightarrow \mathbb{Z}^2$ by $(a,b,c)\rightarrow(b,2a-c)$. This map is onto and its kernel is $\langle (1,0,2) \rangle$. This is because you are in the kernel precisely when $b=0$ and $a,c$ satisfy $2a=c$, clearly any value of $a$ works and $c$ must be $2a$.
You can tell that the quotient can't be $\mathbb{Z}+\mathbb{Z}/2$ because if $(a,b,c)$ has order 2 in the quotient $b=0$ and we get $2(2a)=2b$ which implies $2a=b$ meaning it was already 0 in the quotient.