# (Dis)proving isomorphism between $U= \{ z \in \Bbb C \mid |z| = 1 \}$ and $\Bbb R$

by kanderson8   Last Updated January 12, 2018 20:20 PM

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.

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In $U$ there is an element of order $4$ (that is, $i$). None of the other two groups has one.