Theorem: Correspondence Theorem (Group Theory)

Let $G$ be a group.
Let $N \lhd G$ be a normal subgroup of $G$.
Then every subgroup of the quotient group $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$.
Conversely, if $N \le H \le G$ then $H / N \le G / N$.
The correspondence between subgroups of $G / N$ and subgroups of $G$ containing $N$ is a bijection.
This bijection maps normal subgroups of $G / N$ onto normal subgroups of $G$ which contain $N$.

Proof:
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:
:$h'N \in H'$
From $H' \lhd G / N$:
For all $x \in G$:
:$\paren {x N}^{-1} \paren {h' N} \paren {x N} \in H'$
From $N \lhd G$:
:$x^{-1} h' x N \in H'$
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}}