Finding homomorphism from finite abelian group G

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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?




abstract-algebra group-homomorphism




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  • 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...
    – DonAntonio
    Aug 20 at 10:37










  • @DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:37










  • @Va That is correct.
    – DonAntonio
    Aug 20 at 10:38










  • @DonAntonio How to define a non trivial homomorphism from G to 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:39










  • @VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
    – Santana Afton
    Aug 20 at 11:55














up vote
2
down vote

favorite












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?




abstract-algebra group-homomorphism




share|cite|improve this question




















  • 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...
    – DonAntonio
    Aug 20 at 10:37










  • @DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:37










  • @Va That is correct.
    – DonAntonio
    Aug 20 at 10:38










  • @DonAntonio How to define a non trivial homomorphism from G to 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:39










  • @VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
    – Santana Afton
    Aug 20 at 11:55












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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?




abstract-algebra group-homomorphism




share|cite|improve this question












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?





abstract-algebra group-homomorphism





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 20 at 10:27









Vaibhav Kalia

204




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  • 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...
    – DonAntonio
    Aug 20 at 10:37










  • @DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:37










  • @Va That is correct.
    – DonAntonio
    Aug 20 at 10:38










  • @DonAntonio How to define a non trivial homomorphism from G to 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:39










  • @VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
    – Santana Afton
    Aug 20 at 11:55
















  • 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...
    – DonAntonio
    Aug 20 at 10:37










  • @DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:37










  • @Va That is correct.
    – DonAntonio
    Aug 20 at 10:38










  • @DonAntonio How to define a non trivial homomorphism from G to 1,-1
    – Vaibhav Kalia
    Aug 20 at 10:39










  • @VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
    – Santana Afton
    Aug 20 at 11:55















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...
– DonAntonio
Aug 20 at 10:37




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...
– DonAntonio
Aug 20 at 10:37












@DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
– Vaibhav Kalia
Aug 20 at 10:37




@DonAntonio correct me if I am wrong sir but I think only finite subgroup of R* other than trivial is 1,-1
– Vaibhav Kalia
Aug 20 at 10:37












@Va That is correct.
– DonAntonio
Aug 20 at 10:38




@Va That is correct.
– DonAntonio
Aug 20 at 10:38












@DonAntonio How to define a non trivial homomorphism from G to 1,-1
– Vaibhav Kalia
Aug 20 at 10:39




@DonAntonio How to define a non trivial homomorphism from G to 1,-1
– Vaibhav Kalia
Aug 20 at 10:39












@VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
– Santana Afton
Aug 20 at 11:55




@VaibhavKalia Can you find a homomorphism to $mathbbZ/2mathbbZ$ from some particular groups you are familiar with?
– Santana Afton
Aug 20 at 11:55















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