# Answer to Question #23730 in Abstract Algebra for john.george.milnor

Question #23730

For finite abelian groups G and H, show that RG ∼ RH as R-algebras iff |G| = |H| and |G/G2| = |H/H2|.

Expert's answer

Since

*G*is abelian, Wedderburn’s Theoremgives: R*G**∼*R*×· ··×*R*×*C*×· · ·×*C*.*Suppose there are*s*factors ofR, and*t*factors of C, so that*|G|*=*s*+ 2*t*. Thenumber*s*is the number of 1-dimensional real representations of*G*.This is the number of group homomorphisms from*G*to*{±*1*}**⊆*R*, so*s*=*|G/G*2*|*. Therefore, the isomorphism typeof R*G*(as an R-algebra) is uniquely determined by*|G|*and*|G/G*2*|*.The conclusion follows immediately from this.Need a fast expert's response?

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