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.

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