In what follows, the symbol ×, when applied to subgroups, means the join, or the subgroup generated by the two operands. The symbol ∩ denotes the intersection of two subgroups.
You need to be familiar with the theorem that characterizes the join of two subgroups. It is an important part of this proof.
Let A, S, B, and T be subgroups of G, with S normal in A and T normal in B. We want to show that S×(A∩T) is normal in S×(A∩B), and T×(B∩S) is normal in T×(B∩A). Furthermore, the two quotient groups implied by these two relationships are isomorphic.
Instead of building a map one to the other, we will equate the first factor group with (A∩B) mod (A∩T)×(B∩S). This expression is symmetric with respect to our variables. Swap A for B and S for T and find the very same group. Thus the reasoning below implies a parallel isomorphism from the second quotient onto the same group, and the two quotients are isomorphic to each other.
Start with S×(A∩B) mod S×(A∩T), and work towards our symmetric expression.
Since S is normal in A, restrict to B to show S∩B is normal in A∩B. Similarly, T∩A is normal in B∩A.
Join the two groups S∩B and T∩A, which are both normal in A∩B. Let J = (B∩S)×(A∩T). This is the kernel of the destination factor group. Since J is built from normal subgroups, J is normal in A∩B.
Let H = S×(A∩B). Both factors live in A, so H is a subgroup of A.
We know that S is a normal subgroup of A, and A∩B is a generic subgroup of A. When a normal subgroup joins another subgroup, the composite is well characterized by the aforementioned join theorem. Every element of H is defined by a triple xyz, where z is an element of the intersection (in this case S∩B), x is an element of A∩B that represents a coset of B∩S, and y is an element of S that represents a coset of B∩S. Merge y and z together to form an element of S. Thus H consists of elements xy, where y is in S, and x is a coset of B∩S in A∩B.
With S normal in H, a group homomorphism maps H onto the cosets of S, which are faithfully represented by the cosets of S∩B in A∩B. Again, this was described in the theorem on joining subgroups.
Remember that J was defined as B∩S join A∩T. Thus J is a certain collection of cosets of B∩S, all of them living in A∩B. And J is a normal subgroup. So apply a second homomorphism that has J as its kernel. Call this composite homomorphism f. Now f maps H onto (A∩B)/J.
The kernel of f includes (A∩T)×(B∩S), and all of S. Replace B∩S with S, and the kernel is S×(A∩T).
Let's recap. We have shown that H mod S×(A∩T), which is the same as S×(A∩B) mod S×(A∩T), is isomorphic to (A∩B) mod (A∩T)×(B∩S). This is our destination factor group, symmetric in A and B, S and T. That completes the proof.
If you have to go over this three or four times to get it, you're not the only one. :-)