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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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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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The proof of the second isomorphism theorem given in the Judson book, while correct, is far from ideal. Below we use the terminology of Judson's book.

Theorem (2nd isomorphism theorem): Let $G$ be a group, $N$ a normal subgroup of $G$ and $q \colon G \to G/N$ the canonical homomorphism given by $q(g)=gN$. If $H$ is any subgroup of $G$, then $q$ induces and isomorphism $H/H\cap N \to HN/N$.

Proof: Consider the restriction of $q_{|H} \colon H \to G/N$ of $q$ to $H$. It is a homomorphism, thus its image $q(H)$ is a subgroup of $G/N$ (by Proposition 11.4(3) of Judson). Now by definition of $q$ the image of $q_{|H}$ is $q(H)=\{hN: h \in H\}$, which is precisely $HN/N$ and the kernel $$ \ker(q_{|H}) = \{h \in H: h \in \ker(q)=N\} = H \cap N, $$ and thus by the first isomorphism theorem $H/H\cap N \cong HN/N$.

Note: The facts that $H\cap N$ is normal in $H$ and that $HN$ is a subgroup of $G$ are both consequences of the above proof: the first follows from the fact that it is the kernel of $q_{|H}$ (see Theorem 11.5 of Judson) and the second because $HN = q^{-1}(q(H))$ -- see Judson, Theorem 11.4(3) and (4).

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