(Dis)proving isomorphism between $U= \{ z \in \Bbb C \mid |z| = 1 \}$ and $\Bbb R$ I'm beginning to work through A first course in Abstract Algebra by Fraleigh and Katz. In Chapter 1, Section 4 (Groups), there is the following question:

Let $U$ be a set such that $U= \{ z \in \Bbb C \mid |z| = 1  \}$. Show that the group $\langle U, \cdot \rangle$ is not isomorphic to either $\langle \Bbb R,+ \rangle$ or $\langle \Bbb R^*, \cdot \rangle$. (All groups have a cardinality of $|\Bbb R|$.)

I've done a few problems showing that there does exist an isomorphism between groups, but none that show there doesn't exist one. From the text, I understand that I need to try to identify some structural difference between the groups. I can identify one right off the bat: for all $x \in U$, there exists some $c \in U$ such that $x \cdot x = c$. However, for $\Bbb R^*$, this isn't the case, as $x \cdot x = -1$ has no solution. This shows that the two groups are not isomorphic. I'm struggling, however, to show that $\langle U, \cdot \rangle$ is not isomorphic to $\langle \Bbb R,+ \rangle$. For all $c \in \Bbb R$, we can define $x$ to be $\frac c 2$,  so $x + x = c$ always has a solution. This shows that there is no isomorphism between $\langle \Bbb R,+ \rangle$ and $\langle \Bbb R^*, \cdot \rangle$ but doesn't help with $\langle U, \cdot \rangle$. Both groups have the same cardinality and are abelian, so we can't use order or commutativity to disprove isomorphism. 
On the contrary, I feel like we can define an isomorphism $\phi$: $\langle \Bbb R,+ \rangle \to \langle U, \cdot \rangle$ as $\phi (n) = \zeta^n$, where $\zeta^n$ is the $n^{th}$ root of unity. This has the principle of homomorphism because $\phi (n + m) = \zeta^{n+m} = \zeta^n \zeta^m = \phi(n) \cdot \phi(m)$. Also, since each group has the same cardinality and we know that for every distinct $n \in \Bbb R$ we have a distinct $\zeta^n \in U$, then $\phi$ is bijective. Since it is both bijective and homomorphic, $\phi$ is isomorphic.
I'm sure that my proof is incorrect somewhere but I'm not experienced enough in writing proofs to identify the error, nor can I think of some difference in binary structure between the groups that disproves isomorphism. Any assistance (both in finding my error and helping me solve the problem) would be greatly appreciated.
 A: In $U$ there is an element of order $4$ (that is, $i$). None of the other two groups has an element of order $4$.
A: First, here's an easy way to see $(\mathbb{R}, +)$ and $(U,\cdot)$ are not isomorphic. Every nonzero element of $\mathbb{R}$ has infinite order, but $U$ has many nonzero elements of finite order: consider $-1$ or any root of unity.  (Similarly, $U$ has $2$ elements of order $3$, while $\mathbb{R}^*$ has none.)
One problem with your proposed proof is that, contrary to your claim, $\phi$ is not a homomorphism, since the product of an $n^\text{th}$ and $m^\text{th}$ root of unity need not be an $(n+m)^\text{th}$ root of unity.  For example, consider $n = 2$ and $m = 4$.  Then $\zeta_2 = -1$, $\zeta_4 = i$, but $\zeta_2 \zeta_4 = -i$ is not a sixth root of unity since $(-i)^6 = -1$.  In fact, $\zeta_n \zeta_m = \zeta_{\text{lcm}(m,n)}$ when $m$ and $n$ are positive integers.  (I should also add that I think most people consider it very bad practice to write $n$ for a real number that need not be an integer!)
It's also not clear what you mean by an $r^\text{th}$ root of unity when $r$ is not a positive integer.  Is it just any element $u \in U$ such that $u^r = 1$?  Even for a positive integer $n$, there are $n$ $n^\text{th}$ roots of unity, namely $e^{2 \pi i k/n}$ for $k \in \{0, \ldots, n-1\}$.  Or maybe you mean $\zeta_r = e^{2 \pi i/r}$?  In that case, it's even clearer that the map is not a homomorphism, since that would mean $e^{2 \pi i/r} e^{2 \pi i/s} = e^{2 \pi i/(r+s)}$.  It's also not injective, since $e^{2 \pi i/1} = 1 = e^{2 \pi i/-1}$.
I hope this has helped point out the problems with the definition of your map.
