Prove that the quotient of the following groups $G/H$ is the cyclic group of order 4 This is probably a very basic exercise of group theory.
I am trying to show that $G/H$ is isomorphic to the cyclic group $\mathbb Z_4$ being $G=\mathbb Z\oplus \mathbb Z_2$ and $H$ the (normal) subgroup of elements of the form $(2n,[n])\in  \mathbb Z\oplus \mathbb Z_2$.
I am looking for a Theorem or result that make the computation of $G/H$ relatively easily.
In any case, I am not able to show $G/H\cong \mathbb Z_4$ just by using the definition of quotient group.
Any help would be appreciated
 A: Here is how we can obtain this result if we use only the definitions of the cosets and the quotient group.

*

*We prove that $(1,[0]),(2,[0]),(3,[0])\notin H$ and $(4,[0])=(2\cdot2,[2])\in H$.


*We deduce from here that the 4 cosets $H$, $(1,[0])+H$, $(2,[0])+H$, $(3,[0])+H$ are pairwise distinct.


*Now it is clear that  the quotient group $G/H$ is a cyclic group of order 4 with generating $(1,[0])+H$.


*Finally, we know that every cyclic group of order $m$ is isomorphic to $\mathbb{Z}_m$.
Apparently this was meant in the comments.
A: One presentation for $G=\Bbb Z\oplus \Bbb Z_2$ is
$$P=\langle a,b\mid b^2, ab=ba\rangle.$$
Note that $H$ is given by the normal subgroup $Q=\langle a^2b\rangle$; therefore, the quotient $G/H$ is given by $P/Q$, which is
$$P/Q\cong\langle a,b\mid a^2b, b^2, ab=ba\rangle,$$
so we can write $b=a^{-2}$ to get
$$P/Q\cong \langle a\mid a^2(a^{-2}), (a^{-2})^2, aa^{-2}=a^{-2}a\rangle,$$
which is, in turn, isomorphic to
$$P/Q\cong\langle a\mid a^{-4}\rangle;$$
but if $x=a^{-1}$, then
$$P/Q\cong\langle x\mid x^4\rangle.\tag{1}$$
But the RHS of $(1)$ is a presentation for $\Bbb Z_4$.
