Let $G$ be a abelian group and $H\subseteq G$ is a subgroup. Prove: There exists homomorphism $Q$ such that $\ker Q=H$.

Question

Let $G$ be a abelian group and $H\subseteq G$ is a subgroup. Prove: There exists homomorphism $Q$ such that $\ker Q = H$.

So far I've tried to define homomorphism $Q$ such that $\ker Q = H$ with no success. I'm not even sure how to use the fact that $G$ is abelian.

Any hints will be useful.

Answer 1

Hint : $Q : G \to G/H $

by $Q(g) =g+H$

Answer 2

Since $G$ is abelian, every subgroup of $G$ is normal in $G$ (you can prove this straightly from the definition of normal subgroup). In particular $H$ is normal in $G$, so the quotient $ G / H$ is a group itself. Then as it's also menioned above, define $Q$ in the following way $$ \begin{cases} Q: &G \to G/H \\ &g \mapsto g \cdot H \end{cases} $$ Try to prove that this $Q$ between these two groups is actually a homomorphism between them; And then determine it's kernel.

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Edits (See the discussion below with @Shaun)

Note that if $H$ is an arbitrary subset of $G$, there might not be any homomorphism $Q$ that $ \mathrm{ker} (Q) =H$; this is because every kernel of a homomorphism from $G$ to any other group, must be a subgroup of $G$ (you can also prove this directly). And so the statement of your question is flawed, and $H$ must have assumed to be (at least) a subgroup.