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

  1. \(H\) is a normal subgroup of \(G\) if and only if \(H/N\) is a normal subgroup of \(G/N\).
  2. \((A \cap B)/N = A/N \cap B/N\)
  3. \(A \subseteq B \iff A/N \subseteq B/N\)
  4. \(A \subseteq B \implies [B : A] = [B/N : A/N]\)
  5. \(\langle A, B \rangle/N = \langle A/N, B/N \rangle\)

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 \(\phi : \{H \leq G : N \subseteq H \} \to \{\bar{H} \leq G/N\}\) such that \(\phi(H) = H/N\).

First, note that because \(N \subseteq H \leq G\) for \(N \trianglelefteq G\), we have by this result that \(N \trianglelefteq H\) 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, \(\phi\) is a well defined map.

Now, to prove \(\phi\) is surjective, define for each \(\bar{H} \leq G/N\), a set \(H = \{h \in G: hN \in \bar{H}\}\). 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 \(aN, bN \in \bar{H}\), and consider that

\[ (aN)(bN)^{-1} = (ab^{-1})N \in \bar{H}\]

and thus \(ab^{-1} \in H\).

For injectivity, suppose that \(H\) and \(M\) are distinct subgroups of \(G\) containing \(N\). Without loss of generality, assume there exists an \(h \in H \setminus M\). Assume now by way of contradiction that \(hN \in M/N\), that is, \(hN = mN\) for some \(m \in N\). This however implies that \(hm^{-1} \in N \subseteq H \cap M\), and thus that \(hm^{-1} = m'\) for some \(m' \in M\). This means that \(h = mm'\) and hence \(h \in M\), a contradiction. Therefore \(hN \not\in M/N\) and \(H/N \neq M/N\).