Let $f : G \rightarrow \mathbb{R}\, $ be a homomorphism. Then, I know that $ \ker f$ is a normal closed subgroup in $G$. Why is $G/ \ker f$ abelian?
2026-03-27 14:45:14.1774622714
Why is $G/ \ker f$ abelian?
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The isomorphism theorem says $G/\ker f \cong \operatorname{img} f ⊆ ℝ$ and subgroubs of abelian groups are abelian.