determining the quotient group in Mayer-Vietoris sequence

Question

I am having trouble to determine the quotient group in the following Mayer-Vietoris sequence. I know this problem in Hatcher exists here but my question is not to have a solution (because I do have one). I want to understand how we come up with it.

X is the space obtained by attaching a Mobius strip M to the projective plane $RP_2$ by attaching its boundary to the copy of $S_1= RP_1 \subset RP_2$. I was able to find out using Mayer-vietoris sequence that $H_1(X)=(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}) / Im \phi$ where $\phi(c)=(a,-2b)$. So $H_1(X)=\langle a,b\rangle / \langle 2a,a-2b \rangle $.

Up to here, I understand all the steps. Now this is what I am struggling with: How can we determine $H_1(X)$? I know the answer should be $\mathbb{Z_4}$ but I am having trouble to see that.

PS: I saw in one of the sites at the internet this: $\langle a,b \rangle / \langle 2a,a-2b \rangle =\langle a+2b,b \rangle /\langle a+2b,-4b \rangle \simeq\mathbb{Z_4}$ but I don't understand why at all. I understand that this is an algebraic question but I would rather put the context to it as well.

Answer 1

I don't understand the explanation you've quoted as well. However $a-2b$ in the denominator means means that $[a]=2[b]$ in the quotient. Thus the $a$ generator is redundant and we can replace it with $2b$. This gives us the following:

$$\langle a,b\rangle/\langle 2a,a-2b\rangle\simeq\langle 2b,b\rangle/\langle 4b,0\rangle =\langle b\rangle/\langle 4b\rangle\simeq\mathbb{Z}_4$$

Note that this is not a very precise and formal proof. More like an intuition. The first isomorphism requires a bit of work with free (abelian) groups.