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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 &sim; RH as rings.
1
2013-02-05T10:40:06-0500
We claim that there exists a group Ksuch that K &times; G &sim; K &times; H as groups. First let us assumethis claim. Then
Z[K &times; G] &sim; Z[K &times; H] as rings. It is easy to see that Z[K &times; G] &sim; (Z[K])[G], andsimilarly Z[K &times; H] &sim; (Z[K])[H]. Therefore, for the integral group ring R: = Z[K], we have RG &sim; RH as rings.
To prove our claim, take the group K= (G &times; H) &times; (G &times; H)&times;&middot; &middot; &middot; . We have
G &times; K &sim; G &times; (H &times; G) &times; (H &times; G)&times;&middot; &middot; &middot;
&sim; (G &times; H) &times; (G &times; H)&times;&middot;&middot; &middot; &sim;= K, and
H &times; K &sim;= H &times; (G &times; H) &times; (G&times; H)&times;&middot; &middot; &middot;
&sim; (H &times; G) &times; (H &times; G)&times;&middot;&middot; &middot; &sim;= K.
In particular, G &times; K &sim; H &times; K, as desired.

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