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Here it is the Second Isomorphism Theorem with its proof:

Theorem $\mathbf{11.4}$ (Second Isomorphism Theorem) Let $H$ be a subgroup of a group $G$ (not necessarily normal in $G$) and $N$ a normal subgroup of $G$. Then $HN$ is a subgroup of $G$, $H\cap N$ is a normal subgroup of $H$, and $$H/H\cap N\cong HN/N.$$

$\text{P}\scriptstyle{\text{ROOF}}$. We will first show that $HN=\{hn:h\in H,n\in N\}$ is a subgroup of $G$. Suppose that $h_1n_1,h_2n_2\in HN$. Since $N$ is normal, $(h_2)^{-1}n_1h_2\in N$. So $$(h_1n_1)(h_2n_2)=h_1h_2((h_2)^{-1}n_1h_2)n_2$$ is in $HN$. The inverse of $hn\in HN$ is in $HN$ since $$(hn)^{-1}=n^{-1}h^{-1}=h^{-1}(hn^{-1}h^{-1}).$$ $\quad$ Next, we pove that $H\cap N$ is normal in $H$. Let $h\in H$ and $n\in H\cap N$. Then $h^{-1}nh\in H$ since each element is in $H$. Also, $h^{-1}nh\in N$ since $N$ is normal in $G$; therefore, $h^{-1}nh\in H\cap N$.
$\quad$ Now 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 $$\ker\phi=\{h\in H:h\in N \}=H\cap N,$$ $HN/N=\phi(H)\cong H/H\cap N$. $\tag*{$\square$}$

I don't understand why $\phi(H) \cong H/ \ker\phi$ (in the last paragraph). It says according to the First Isomorphism Theorem, but I couldn't find it in there as here it is the theorem with its proof:

Theorem $\mathbf{11.3}$ (First Isomorphism Theorem) If $\psi:G\to H$ is a group homomorphism with $K=\ker\psi$, then $K$ is normal in $G$. Let $\phi:G\to G/K$ be the canonical homomorphism. Then there exists a unique isomorphism $\eta:G/K\to\psi(G)$ such that $\psi=\eta\phi$.

$\text{P}\scriptstyle{\text{ROOF}}$. We already know that $K$ is normal in $G$. Define $\eta:G/K\to\psi(G)$ by $\eta(gK)=\psi(g)$. We first show that $\eta$ is a well-defined map. If $g_1K=g_2K$, then for some $k\in K$, $g_1k=g_2$; consequently, $$\eta(g_1K)=\psi(g_1)=\psi(g_1)\psi(k)=\psi(g_1k)=\psi(g_2)=\eta(g_2K).$$ Thus, $\eta$ does not depend on the choice of coset representatives and the map $\eta:G/K\to\psi(G)$ is uniquely defined since $\psi=\eta\phi$. We must also show that $\eta$ is a homomorphism, but $$\begin{align}\eta(g_1Kg_2K)&=\eta(g_1g_2K)\\&=\psi(g_1g_2)\\&=\psi(g_1)\psi(g_2)\\&=\eta(g_1K)\eta(g_2K).\end{align}$$ Clearly, $\eta$ is onto $\psi(G)$. To show that $\eta$ is one-to-one, suppose that $\eta(g_1K)=\eta(g_2K)$. Then $\psi(g_1)=\psi(g_2)$. This implies that $\psi(g_1^{-1}g_2)=e$, or $g_1^{-1}g_2$ is in the kernel of $\psi$; hence, $g_1^{-1}g_2K=K$; that is, $g_1K=g_2K$. $\tag*{$\square$}$

Detailed and easy to understand explanation of why $\phi(H) \cong H/ \ker\phi$ in the Second Isomorphism Theorem would really be appreciated.

Edit - also there is no such $HN/N$ as the co-domain in the first theorem.

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  • $\begingroup$ It says it in the statement of the first theorem. $\endgroup$ – Euler....IS_ALIVE Aug 7 '15 at 3:31
  • $\begingroup$ They are just using the statement of first isomorphism theorem nothing else. $\endgroup$ – Chiranjeev_Kumar Aug 7 '15 at 3:33
  • $\begingroup$ @Euler....IS_ALIVE yes I know and I've uploaded the theorem and proof. many things happens in the first theorem and no clear finding my question in it. $\endgroup$ – L.G. Aug 7 '15 at 3:34
  • $\begingroup$ @Chiranjeev would you please tell me where in the first theorem $\phi(H) \cong H/ \text{ker {$\phi$}}$ or proof of it independent of the first theorem? Thank you $\endgroup$ – L.G. Aug 7 '15 at 3:36
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    $\begingroup$ In the context of the first isomorphism theorem statement, replace $G$ with $H$, $K$ with $ker{\phi}$ and $\psi(G)$ with $\phi (H)$. $\endgroup$ – Taylor Aug 7 '15 at 3:40
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The First Isomorphism Theorem states that for groups $G, H$ and a homomorphism $\phi : G \to H$, $G/ker(\phi) \cong \phi(G)$.

In the Second Isomorphism Theorem, we have the groups $H$ and $HN/N$. As $N$ is normal, we have $HN/N$ is a quotient group. The natural projection map $\phi : H \to HN/N$ sending $h \mapsto hN$ is a homomorphism. Thus, $\phi(H) \cong H/ker(\phi)$ by the First Isomorphism Theorem.

It sounds like your confusion is to the correctness of the First Isomorphism Theorem. The intuition is that the interaction of the elements of $\phi(G)$ is equivalent to the interaction of the fibers of the homomorphism. The kernel of the homomorphism acts on the cosets.

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From 11.3, if $\psi$ is a homomorphism from G to H then $G/ker(\psi)$ is isomorphic to $\psi(G)$. In that sentence, change $G$ to $H$,change $\psi$ to $\phi$,and change $H$ to $HN/N$.

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