Correspondence Theorem

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle x, y \in \beta \left({H'}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle x N, y N \in H'\)	\(\displaystyle \)	by definition of $\beta$
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \left({x N}\right) \left({y N}\right) = x y N \in H'\)	\(\displaystyle \)	as $N$ is normal
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle x y \in \beta \left({H'}\right)\)	\(\displaystyle \)	by definition of $\beta$

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

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 $\alpha \left({H}\right) \le G / N$.

• 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:

	\(\displaystyle \)	\(\displaystyle \left({\beta \circ \alpha}\right) \left({H}\right)\)	\(=\)	\(\displaystyle \beta \left({H / N}\right)\)	\(\displaystyle \)	by definition of $\alpha$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: g N \in H / N}\right\}\)	\(\displaystyle \)	by definition of $\beta$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle H\)	\(\displaystyle \)	???

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

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

	\(\displaystyle \)	\(\displaystyle \left({\alpha \circ \beta}\right) \left({H'}\right)\)	\(=\)	\(\displaystyle \alpha \left({\left\{ {g}\right\} \in G: g N \in H'}\right)\)	\(\displaystyle \)	by definition of $\beta$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g N \in H'}\right\}\)	\(\displaystyle \)	by definition of $\alpha$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle H'\)	\(\displaystyle \)	???

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 $G$.

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 47 \eta$
• John F. Humphreys: A Course in Group Theory (1996): $\S 7$: Theorem $7.14$