Correspondence Theorem

The correspondence theorem in group theory gives a classification of all subgroups of a quotient group in terms of subgroups of the group that it is a quotient of.

Theorem

Given a group $G$ with a normal subgroup $N$ , there is a bijection between the set of subgroups of $G$ containing $N$ and the subgroups of $G / N$

{H \leq G : N \subseteq H} \to {\bar{H} \leq G / N}

given by

H \mapsto H / N .

This gives us a full classification of all the subgroups of the quotient group $G / N$ . In particular they are of the form $H / N$ where $N \subseteq H \leq G$ .

Theorem

The correspondence above satisfies

$H$ is a normal subgroup of $G$ if and only if $H / N$ is a normal subgroup of $G / N$ .
$(A \cap B) / N = A / N \cap B / N$
$A \subseteq B ⟺ A / N \subseteq B / N$
$A \subseteq B ⟹ [B : A] = [B / N : A / N]$
$⟨ A, B ⟩ / N = ⟨ A / N, B / N ⟩$

This list is far from exhaustive; a lot of the structure of the subgroups of the initial group are preserved in the quotient.

Intuitively, consider $H$ a subgroup which contains $N$ but is containing in $G$ , and this subgroup corresponds with the set of cosets of $N$ inside it.

In the following diagram, we have $G$ on the left partitioned into cosets of $N$ , with $H$ being a subgroup containing some number of those cosets. The way $H$ nicely contains an integer number of cosets is a consequence of the first result we prove below.

The set of elements in $H$ , the green rectangle on the left, corresponds with the set of cosets of $H$ in $N$ , marked in green on the right.

This correspondence means that our bijection maps $H$ to $H / N$ , and hence similarly it would map different selections of $H$ to the corresponding $H / N$ as follows:

This point is made simply to make it clear that the mapping is not a correspondence between $H$ and $H / N$ (in terms of their elements). It's a correspondence between all such subgroups $H$ and $H / N$ .

We now proof the first result.

Proof

Denote the bijection by $ϕ : {H \leq G : N \subseteq H} \to {\bar{H} \leq G / N}$ such that $ϕ (H) = H / N$ .

First, note that because $N \subseteq H \leq G$ for $N ⊴ G$ , we have by this result that $N ⊴ H$ and hence $H / N$ is a well defined group. Furthermore, because $H / N \subseteq G / N$ , and these groups use the same operation, $H / N$ is indeed a subgroup of $G / N$ . As such, $ϕ$ is a well defined map.

Now, to prove $ϕ$ is surjective, define for each $\bar{H} \leq G / N$ , a set $H = {h \in G : h N \in \bar{H}}$ . We will prove $H$ is a subgroup of $G$ using the subgroup tests given ${0} \neq H \subseteq G$ . Let $a, b \in H$ , hence $a N, b N \in \bar{H}$ , and consider that

(a N) (b N)^{- 1} = (a b^{- 1}) N \in \bar{H}

and thus $a b^{- 1} \in H$ .

For injectivity, suppose that $H$ and $M$ are distinct subgroups of $G$ containing $N$ . Without loss of generality, assume there exists an $h \in H ∖ M$ . Assume now by way of contradiction that $h N \in M / N$ , that is, $h N = m N$ for some $m \in N$ . This however implies that $h m^{- 1} \in N \subseteq H \cap M$ , and thus that $h m^{- 1} = m^{'}$ for some $m^{'} \in M$ . This means that $h = m m^{'}$ and hence $h \in M$ , a contradiction. Therefore $h N \notin M / N$ and $H / N \neq M / N$ .