Question #23731

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

Expert's answer

We claim that there exists a group *K*such 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.

Z[

To prove our claim, take the group

In particular,

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