# (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.

• In $U$ you have an element $i\neq 1$ such that $i^4=1$ and $i^2\neq 1$. But in $\mathbb{R}$ with $+$ adding a non-zero element to itself many times never gives $0$. The same happens in $\mathbb{R}^*$. Multiplying an element that is not $1$ to itself many times doesn't give you $1$ until the $4$-th time. Jan 12 '18 at 20:21
• This is definitely outside my expertise, but I’m still curious: what is $\Bbb{R}^*$? Jan 13 '18 at 0:09

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

• Thanks, this convinces me that there doesn't exist an isomorphism between the groups, but I still don't understand what the error in my proof is. If you find time to add something about why my proposed isomorphism is invalid, I'd be happy to accept this answer. Jan 12 '18 at 20:54
• Not even sure where to start, it is quite broken, actually... First: how do you even define your isomorphism? At some points in your proof it seems that $\zeta$ depends on $n$, in others it seems that $\zeta$ is chosen independently of any $n$. Also, you mention "roots of unity" - how do you define e.g. $\pi$'th root of unity? (Note: $n\in\mathbb R$). Finally, you merely claim that $\phi$ is injective - I don't see the proof, and subject to other constraints I cannot imagine you can have any.
– user491874
Jan 12 '18 at 21:08
• I suspect, after clearing up all the previous doubts, you will still be basically claiming that $\phi(n)=\zeta^n$, but if $\zeta$ is the $n$th root of unity in any conceivable sense, then $\zeta^n=1$, so $\phi=1$ - a trivial homomorphism which is not isomorphism. So, sorry but I cannot make a whole lot of sense of your proof otherwise.
– user491874
Jan 12 '18 at 21:10
• @kanderson8 I am putting this into comments, but if you can clarify your proof and I have something valuable to add, I will update the actual answer.
– user491874
Jan 12 '18 at 21:12
• @kanderson8 Oh, I forgot, the argument about cardinality does not work for infinite sets... You can have two sets of the same cardinality, one properly contained in the other one (e.g. even integers as a proper subset of all integers).
– user491874
Jan 12 '18 at 21:15

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.