I was computing the homology group of a 2-complex with $\mathbb{Z}$ coefficients. Doing so, I obtained $$ H_1 = \frac{\langle a,b,c,d \rangle}{\langle 2a+2b, 2a+2b+c+d\rangle} $$ I then tried to simplify the expression for $H_1$ in two different ways. Unfortunately, I obtained two different answers. So at most one (if any) can be correct. I was hoping someone could explain where I went wrong and why my step is incorrect.
First Method: $$ H_1 = \frac{\langle a,b,c,d \rangle}{\langle 2a+2b, 2a+2b+c+d\rangle} = \frac{\langle a,b,c,d \rangle}{\langle a = -b, c=-d\rangle} \cong \mathbb{Z}^2 $$
Second Method: $$ H_1 = \frac{\langle a,b,c,d \rangle}{\langle 2a+2b, 2a+2b+c+d\rangle} \cong \frac{\langle a,a+b,c,c+d \rangle}{\langle 2(a+b), c+d \rangle} \cong \mathbb{Z}_2 \times \mathbb{Z}^2 $$
What element of $\mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0\rightarrow a = -b$? I'm not familiar with that element of $\mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)
It might be better to think of this as row reduction, using only integer multiplication, of $$ \begin{pmatrix} 2 & 2 & 0 & 0 \\ 2 & 2 & 1 & 1 \end{pmatrix} $$ to get $\langle 2a + 2b, 2a+2b+c+d\rangle = \langle 2(a+b), c+d \rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis $$ (a,b,c,d) \rightarrow (a,a+b,c,c+d) $$ to simplify finding your quotient).