In my book it gives the example of a quotient group $\mathbb{R}^2/\mathbb{Z}^2$ and you have to find all the elements of order 2.
so if we take $\overline{(x,y)}$ an element of the above quotient group then $\overline{(2x,2y)}=0$
Now it says $2x= ...,-1, 0, 1, 2, 3,....$ and same for $2y$. Now shouldnt it be $2x=0,1$ because $\overline{(x,y)}$ is between 0 and 1. This is the method they give for the example in $\mathbb{R}/\mathbb{Z}$ so i am confused on how that is changed for $\mathbb{R}^2/\mathbb{Z}^2$.
Observe that $\;\overline{(x,y)}=(x,y)+\Bbb Z^2\;$ has order two iff $$\;2(x,y)\in\Bbb Z^2\iff (2x,2y)\in\Bbb Z^2\iff x,y\in\frac12\Bbb Z$$
i.e.: both $\;x,y\;$ are rational half integers (like $\;\frac12,\,\frac{-17}2\;,\;\;\frac{51}2,\,\text{etc.}\;$) .
Suppose now that we have two such elements
$$\;\overline{(x,y)}\;,\;\overline{(a,b)}\;,\;\;a,b,x,y\in\frac12\Bbb Z\,,\,\,\text{say}\;\;a=\frac\alpha2,\,b=\frac\beta2,\,x=\frac\xi2,\,y=\frac\upsilon2$$
then we have that
$$\overline{(a,b)}=\overline{(x,y)}\iff(a-x,\,b-y)\in\Bbb Z^2\iff\frac{\alpha-\xi}2\,,\,\frac{\beta-\upsilon}2\in\Bbb Z\iff$$
$$\iff \alpha-\xi=0\pmod 2\;,\;\;\text{and also}\;\;\beta-\upsilon=0\pmod2$$
and this means $\;\alpha,\,\xi\;$ have the same parity, and $\;\beta,\,\upsilon\;$ have the same parity.
Try now to take it from here...