(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.



Answers 1


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

user8734617
user8734617
January 12, 2018 20:19 PM