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The Second Isomorphism Theorem: Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then $$H/(H\cap N)\cong(HN)/N$$

There is the proof of Abstract Algebra Thomas by W. Judson:

Define a map $\phi$ from $H$ to $HN/N$ by $H\mapsto hN$. The map $\phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $\phi$ is a homomorphism because $$\phi(hh')=hh'N=hNh'N=\phi(h)\phi(h')$$ By the First Isomorphism Theorem, the image of $\phi$ is isomorphic to $H/\ker\phi$, that is $$HN/N=\phi(H)\cong H/\ker\phi$$ Since $$\ker\phi=\{h\in H:h\in N\}=H\cap N$$ $HN/N=\phi(H)\cong H/H\cap N$

My question:

Is it necessary to prove that the map $\phi$ is onto? Can we only prove that $\phi$ is well defined and the image of $\phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.

Thank you.

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1 Answer 1

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The First Isomorphism Theorem states that if $\varphi: G \to G'$, then $\mathrm{im}(\varphi) \cong G/\mathrm{ker}(\varphi)$. If we do not know that your $\phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H \cap N \cong \mathrm{im}(\phi) \subseteq HN/N$, which does not finish the job.

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