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If $\phi$ is a morphism between groups $G$ and $H$, is $G$ isomorphic to $$ker(\phi)\times im(\phi)$$ ? Why ? Thanks.

Asinus
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  • It's not true for modules either: $\mathbb{Z}$-modules are exactly abelian groups, for example. It's true for vector spaces (modules over a field), though. – Najib Idrissi May 28 '14 at 12:11

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No. Consider $\phi : \mathbb{Z}_4 \to \mathbb{Z}_2$, $a \mapsto 2a$. We have $\ker\phi = \operatorname{im}\phi = \mathbb{Z}_2$, but $\mathbb{Z}_4 \not\cong \mathbb{Z}_2\times\mathbb{Z}_2$.

What is true however is that $G/\ker\phi \cong \operatorname{im}\phi$. This is known as the First Isomorphism Theorem which holds for both groups and modules.

Michael Albanese
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  • thanks. then There is something i don't understand in a proof. I'm going to ask it as another question. – Asinus May 28 '14 at 12:22