Correspondence Theorem (Group Theory)

Theorem
Let $$G$$ be a group.

Let $$N$$ be a normal subgroup of $$G$$.

Then every subgroup of the quotient group $$G / N$$ is of the form $$H / N$$, 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 $$\left\{{h N}\right\}$$ of left cosets of $$N$$ in $$G$$.

Let us define the subset $$\beta \left({H'}\right) \subseteq G$$:


 * $$\beta \left({H'}\right) = \left\{{g \in G: g N \in H'}\right\}$$

Then clearly $$N \subseteq \beta \left({H'}\right)$$. Also:


 * $$e_G \in N$$, so $$e_G \in \beta \left({H'}\right)$$.


 * Let $$x, y \in \beta \left({H'}\right)$$. Then:

$$ $$ $$ $$


 * $$\left({x N}\right)^{-1} = x^{-1} N \implies x^{-1} \in \beta \left({H'}\right)$$.

Thus, by the Two-step Subgroup Test, $$\beta \left({H'}\right) \le G$$ that contains $$N$$.


 * Conversely, let $$H$$ be such that $$N \le H \le G$$.

Let $$\alpha \left({H}\right) = \left\{{h N: h \in H}\right\} \subseteq G / N$$.

It is easily checked that $$\alpha \left({H}\right) \le G$$.


 * 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 $$H \triangleleft G$$ such that $$N \le H$$.

We show that $$\alpha \left({H}\right) \triangleleft G / N$$. This follows because:

$$\left({g N}\right) \left({h N}\right) \left({g N}\right)^{-1} = g h g^{-1} N \in H N$$

Conversely, let $$H' \triangleleft G / N$$.

It is easily checked that $$\beta \left({h'}\right) = \left\{{g \in G: g N \in h'}\right\}$$ is a normal subgroup of $$G$$.