2.54:

A, B, and C are normal subgroups of G, satisfying AC = BC and A ^ C = B ^ C (^ means intersection), and A is a subgroup of B. Click here to see a diagram. We are to show A=B. Since A is a subset of B, A is contained in the kernel of the canonical homomorphism AC ---> AC/B = BC/B, hence there is a well-defined homomorphism

f : AC/A ---> AC/B = BC/B.

The kernel of f is equal to B/A. (f is defined by f(xA) = xB.) By the second isomorphism theorem, there are isomorphisms

g : C/(A ^ C) ---> AC/A

and o

h : C/(B ^ C) ---> BC/B

induced by the inclusions C ----> AC and C ----> BC. Then the composite h^-1 o f o g : C/(A ^ C) ---> C/(B ^ C) = C/(A ^ C)

is the identity map. Here is a diagram:

C/(A ^ C) ---> AC/A ---> AC/B = BC/B ----> C/(B ^ C) = C/(A ^ C),

with f in the middle, and the two outer maps being insomorphisms. It follows that f is injective, so the kernel B/A of f is trivial, hence A=B. (This argument is easy to see from the diagram.)