Is $\mathbb{C}^*$ modulo the origins of unity isomorphic to $\mathbb{R}^+$?

A pupil involved me revealing an inquiry from his test in standard group theory, in which they are asked to confirm that $\mathbb{C}^*$ modulo the subgroup of origins of unity is isomorphic to $\mathbb{R}^+$ (in both instances we suggest the multiplicative teams).

Currently this appears to me to be a straightforward mistake in the inquiry. I think they suggested to ask to confirm that $\mathbb{C}^*$ modulo all the components of outright value 1 is isomorphic to $\mathbb{R}^+$ which is really simple to confirm (take the homomorphism mapping $z$ to $|z|$ and also make use of the first homomorphism theory). Nonetheless, I could not confirm the case concerning the origins of unity is incorrect ; exists a very easy means to show this?

0
2019-12-02 02:50:43
Source Share
For the purpose of efficiency : $\mathbb{C}^{\ast}$ is isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Z}$ in the noticeable means, and also $\mathbb{C}^{\ast}$ modulo the origins of unity is after that isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Q}$. As a $\mathbb{Q}$ - vector room this is abstractly isomorphic to $\mathbb{R}^{+}$, yet the building and construction of such an isomorphism is most likely to call for the axiom of selection.