Correspondence Theorem (Group Theory)

Theorem
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:

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:

Thus $\beta \circ \alpha = I_X$.

Now let $H' \le G / N$. Then:

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:

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:
 * $\map \beta {H'} \lhd G$