# determining the quotient group in Mayer-Vietoris sequence

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.

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