Let $G$ be a finite abelian group, and let $K$ be a subgroup of $G$. Does $G$ necessarily have a subgroup $H$ such that $H\cong G/K$ and $H\cap K=\langle 0\rangle$?
I'm not sure where to start.
Let $G$ be a finite abelian group, and let $K$ be a subgroup of $G$. Does $G$ necessarily have a subgroup $H$ such that $H\cong G/K$ and $H\cap K=\langle 0\rangle$?
I'm not sure where to start.
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Let $G = \mathbb{Z}_4$ be the cyclic group of order $4$, and $K = \langle 2 \rangle = \{0,2\}$. If such an $H$ existed, $|H| = 2$; but this isn't possible given the condition $H \cap K = \{0\}$.