0
$\begingroup$

For example: $G_1=(R,+) \quad G_2=(R_+,x)$ So the first group are the Real Number closed under addition and the second group are the positive Real Numbers closed under multiplication. My Lector said that the Isomorphism was $f(x)=e^x$ because its a bijection and $e^(x+y)=e^xe^y$. However, if the 2 Groups were not closed under addition and multiplication would the Groups still be isomorphic? (Btw: I did my research but I didnt really know what I should search for.)

$\endgroup$
  • 3
    $\begingroup$ A group is always closed under its operation. $\endgroup$ – Jorge Fernández Hidalgo Jul 9 '15 at 13:42
  • $\begingroup$ @dREaM I mean if in this case the Real Numbers werent closed under addition but, for example, multiplication, so: $G_1=(R,x) \quad G_2=(R_+,x)$. Would there still be the Groupisomorphism between the two Groups? $\endgroup$ – antoni Jul 9 '15 at 13:48
  • $\begingroup$ No, because the real with multipication has element of order $2$. $\endgroup$ – Ofir Schnabel Jul 9 '15 at 13:50
  • 2
    $\begingroup$ @DavidBeck: $(\mathbb{R}, \times)$ is not a group. $\endgroup$ – Michael Albanese Jul 9 '15 at 13:56
  • $\begingroup$ Anyway the answer to the question is yes, the group operation does matter. $\endgroup$ – Derek Holt Jul 9 '15 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.