I'm confused on an abstract math question.
Let $G$ be a finite abelian group, and let $K$ be a subgroup of $G$. Does $G$ necessarily have a subgroup H such that $H≅G/K$ and $H∩K=⟨0⟩$.
I think it is related to Schur-Zassenhaus lemma. I know that If K is a normal subgroup of a finite group G whose order is prime to its index then K has a complement in G. Not sure how to go from here. Any help would be appreciative.
As $G$ is abelian, $K$ is normal in $G$ and you have then a surjective group homomorphism $$\pi:G\rightarrow G/K$$ whose kernel is $K$.
If $H$ is a subgroup of $G$, the restriction of $\pi$ gives rise to a surjective group homomorphism $$\pi':H\to \pi(H).$$ The fact that $H\cap K=\{0\}$ is equivalent to say that $\pi'$ is injective. If moreover you want $H$ to be isomorphic to $G/K$, then $\pi'$ needs to be also surjective (at least when the isomorphism is given by $\pi'$). So what you are looking for is a"section" of $\pi$, i.e. a group homomorphism $s:G/K\rightarrow G$ such that $\pi\circ s$ is the identity of $G/K$ (and then the map $s\circ \pi$ is not necessarily the identity on $G$, but is the identity on $s(G/K)$).
As explained by Cass, there is not always a section to $\pi$, and the easiest case is $\pi\colon \mathbb{Z}/4\rightarrow \mathbb{Z}/2$. In practice, given explicitely $\pi$ you can easily find if it admits a section of not (at least for finite groups).