Problem
I am finding it hard to show that the quotient space $$\{ z_1a + z_2b+z_3c \ \big| \ z_1,z_2,z_3 \in \mathbb{Z} \} / \{ x_1 a + x_2 b – x_2 c \ \big| \ x_1, x_2 \in \mathbb{Z} \}$$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z} / 2\mathbb{Z}$. Two classes of cycles $$ [z] = [z_1 a + z_2 b + z_3 c] $$ and $$ [y] = [y_1 a + y_2 b + y_3 c] $$ are equivalent here if and only if \begin{align*} y – z & \in \{ x_1 a + x_2 b – x_2 c \ \big| \ x_1, x_2 \in \mathbb{Z} \} \\ & \iff y_3 = z_2 + z_3 – y_2. \end{align*} I don't know how to deduce that this determines a $\mathbb{Z} \oplus \mathbb{Z} / 2\mathbb{Z}$ group. Any help here is appreciated.
Background
I am trying to compute the first integral homology group of the Klein bottle, and refer to Ted Shifrin's answer to the following question: Homology groups of the Klein bottle.
The Klein bottle (K) admits a simplicial decomposition according to the following fundamental polygon representation:
Now the relevant chain complexes are $$C_1= \mathrm{span}\{ a,b,c \} = \{ z_1a + z_2b+z_3c \ \big| z_1,z_2,z_3 \in \mathbb{Z} \}$$ and $$C_2 = \mathrm{span}\{ U,L \} = \{ y_1U + y_2L \ \big| y_1,y_2 \in \mathbb{Z} \}.$$
The boundary maps are as follows: $$ \partial_1 (a) = \partial_1 (b) = \partial_1 (c) = 0, \quad \partial_2(U) = a+b-c, \quad \partial_2(L) = a-b+c. $$
\begin{align*} \implies \mathrm{ker}(\partial_1) &= C_1 = \{ z_1a + z_2b+z_3c \ \big| z_1,z_2,z_3 \in \mathbb{Z} \} \cong \mathbb{Z}^3, \\ \mathrm{im}(\partial_2) &= \partial_2 (y_1U + y_2L) \\ &= y_1(a+b-c) + y_2(a-b+c) \\ &= (y_1+y_2)a + (y_1 – y_2)b + (y_2 – y_1)c \\ &= x_1 a + x_2 b – x_2 c, \quad \text{where} \ x_1 = y_1 + y_2, \ x_2 = y_1 – y_2. \end{align*}
\begin{align*} \implies H_1(K) &= \{ z_1a + z_2b+z_3c \ \big| \ z_1,z_2,z_3 \in \mathbb{Z} \} / \{ x_1 a + x_2 b – x_2 c \ \big| \ x_1, x_2 \in \mathbb{Z} \}. \end{align*}
Best Answer
It's easier to see via a change of bases. Your nontrivial boundary map has the matrix representation $\begin{pmatrix}1 & 1\\ 1 & -1 \\-1 & 1\end{pmatrix}$. Row- and column-reducing over $\mathbb{Z}$, we see that this is equivalent to $\begin{pmatrix}1 & 0\\ 0 & 2 \\0 & 0\end{pmatrix}$. Therefore, with respect to these new bases, $\partial_2(x,y) = (x,2y,0)$. This means that in the quotient $H_1(K) \cong \mathbb{Z}^3/\operatorname{im}(\partial_2)$, the first $\mathbb{Z}$-component gets killed off, and the second gets halved. All that is left is $0 \oplus \mathbb{Z}/2 \oplus \mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}/2$.