Let $H'$ be a subgroup of $G / N$, so that it consists of a certain set $\set {h N}$ of left cosets of $N$ in $G$. Let us define the subset $\map \beta {H'} \subseteq G$: :$\map \beta {H'} = \set {g \in G: g N \in H'}$ Then clearly: :$N \subseteq \map \beta {H'}$ Also: :$e_G \in N$ so: :$e_G \in \map \beta {H'}$ Let $x, y \in \map \beta {H'}$. Then: {{begin-eqn}} {{eqn | l = x, y       | o = \in       | r = \map \beta {H'}       | c =  }} {{eqn | ll= \leadsto       | l = x N, y N       | o = \in       | r = H'       | c = Definition of $\beta$ }} {{eqn | ll= \leadsto       | l = \paren {x N} \paren {y N}       | r = x y N \in H'       | c = {{Defof|Quotient Group}}: $G / N$ and as $H'$ is a subgroup of $G / N$ }} {{eqn | ll= \leadsto       | l = x y       | o = \in       | r = \map \beta {H'}       | c = Definition of $\beta$ }} {{end-eqn}} We also have, from Quotient Group is Group: :$\paren {x N}^{-1} = x^{-1} N \implies x^{-1} \in \map \beta {H'}$ Thus, by the Two-Step Subgroup Test, $\map \beta {H'} \le G$ that contains $N$. Conversely, let $H$ be such that $N \le H \le G$. Let $\map \alpha H = \set {h N: h \in H} \subseteq G / N$. It is easily checked that $\map \alpha H \le G / N$. Now, let $X$ be the set of subgroups of $G$ containing $N$ and $Y$ be the set of all subgroups of $G / N$. We now need to show that $\alpha: X \to Y$ is a bijection. We do this by checking that $\beta: Y \to X$ is the inverse of $\alpha$. To do this, we show that $\alpha \circ \beta = I_Y$ and $\beta \circ \alpha = I_X$. Suppose $N \le H \le G$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {\beta \circ \alpha} } H       | r = \map \beta {H / N}       | c = Definition of $\alpha$ }} {{eqn | r = \set {g \in G: g N \in H / N}       | c = Definition of $\beta$ }} {{eqn | r = H       | c = {{Defof|Coset}} $H / N$ }} {{end-eqn}} Thus $\beta \circ \alpha = I_X$. Now let $H' \le G / N$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {\alpha \circ \beta} } {H'}       | r = \map \alpha {\set {g \in G: g N \in H'} }       | c = Definition of $\beta$ }} {{eqn | r = \set {g N \in H'}       | c = Definition of $\alpha$ }} {{eqn | r = H'       | c = Definition of $H'$ }} {{end-eqn}} Thus $\alpha \circ \beta = I_Y$. So, by Bijection iff Inverse is Bijection, $\alpha$ is a bijection. Now let $H \lhd G$ such that $N \le H$. We show that $\map \alpha H = H / N \lhd G / N$. This follows by definition 3 of Normal Subgroup because: for any $h \in H, g \in G$ :$\paren {g N} \paren {h N} \paren {g N}^{-1} = g h g^{-1} N \in H / N$ :$\paren {g N}^{-1} \paren {h N} \paren {g N} = g^{-1} h g N \in H / N$ Conversely, let $H' \lhd G / N$. Recall: :$\map \beta {H'} = \set {g \in G : g N \in H'}$ Hence, for any $x \in G$ we have: {{begin-eqn}} {{eqn | l = x \map \beta {H'} x^{-1}       | r = \set {x g x^{-1} \in G: g N \in H'}       | c =  }} {{eqn | r = \set {y \in G: x^{-1} y x N \in H'}       | c =  }} {{end-eqn}} Now for any $h' \in \map \beta {H'}$, we have: :$h'N \in H'$ For all $x \in G$: :$\paren {x^{-1} N} \paren {h' N} \paren {x N} \in H'$ From $H' \lhd G / N$: :$x^{-1} h' x N \in H'$ This implies: :$h' \in x \map \beta {H'} x^{-1}$ or: :$\map \beta {H'} \subseteq x \map \beta {H'} x^{-1}$ Similarly, we can also show: :$\map \beta {H'} \subseteq x^{-1} \map \beta {H'} x$ Hence by definition 4 of Normal Subgroup: :$\map \beta {H'} \lhd G$ {{qed}}