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

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

Z[

To prove our claim, take the group

In particular,

*K*such that*K × G**∼**K × H*as groups. First let us assumethis claim. ThenZ[

*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.Need a fast expert's response?

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