Answer to Question #23731 in Abstract Algebra for john.george.milnor

Question

Show that, for any two groups G,H, there exists a (nonzero) ring R such that RG ∼ RH as rings.

Answer 1

We claim that there exists a group Ksuch that K × G ∼ K × H as groups. First let us assumethis claim. Then
Z[K × G] ∼ Z[K × H] as rings. It is easy to see that Z[K × G] ∼ (Z[K])[G], andsimilarly Z[K × H] ∼ (Z[K])[H]. Therefore, for the integral group ring R: = Z[K], we have RG ∼ RH as rings.
To prove our claim, take the group K= (G × H) × (G × H)×· · · . We have
G × K ∼ G × (H × G) × (H × G)×· · ·
∼ (G × H) × (G × H)×·· · ∼= K, and
H × K ∼= H × (G × H) × (G× H)×· · ·
∼ (H × G) × (H × G)×·· · ∼= K.
In particular, G × K ∼ H × K, as desired.