Does there exists a homomorphism from a finite abelian group G to R*(Set of reals under multiplication) ?

My try: Suppose f be non trivial homomorphism from G to R*. Since f(G) is subgroup of R* and G is finite implies R* has a finite subgroup which is other than non trivial. Now R* has only finite subgroup {1,-1} this implies f(G)= {1,-1} .Also from property of homomorphism order of f(G) divides order of G this will conclude that order of G is even. Now I am stuck after this.How to proceed?

  • $\begingroup$ That's pretty much there is: for a non-trivial homomorphism to exists, the group has to have a subgroup of index two and thus it has even order... $\endgroup$ – DonAntonio Aug 20 '18 at 10:37
  • $\begingroup$ @DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is {1,-1} $\endgroup$ – Vaibhav Kalia Aug 20 '18 at 10:37
  • $\begingroup$ @Va That is correct. $\endgroup$ – DonAntonio Aug 20 '18 at 10:38
  • $\begingroup$ @DonAntonio How to define a non trivial homomorphism from G to {1,-1} $\endgroup$ – Vaibhav Kalia Aug 20 '18 at 10:39
  • $\begingroup$ @VaibhavKalia Can you find a homomorphism to $\mathbb{Z}/2\mathbb{Z}$ from some particular groups you are familiar with? $\endgroup$ – Santana Afton Aug 20 '18 at 11:55

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