1

Consider the abelian group $U_n=\{a\in \mathbb{Z}_n:(a,n)=1\}$. Is there a natural way to understand it as a subgroup of any other interpretation of the cyclic group of order $n$. For example, consider the group of $n$th roots of unity which is also isomorphic to $\mathbb{Z}_n$. Now in this case, if I define $U_n$ as $\{e^{\frac{2\pi ik}{n}}:(k,n)=1\}$ I am unable to understand a "geometric feel" of the elements of $U_n$. Is it possible to do so for any other interpretation of the cyclic group of order $n$? Is there an interpretation devoid of any reference to $\mathbb{Z}_n$?

  • 4
    $U_n$ can be considered to be the group of authomorphisms of $(\mathbb Z_n,+)$. – Thomas Andrews May 10 '13 at 15:35
  • I don't think $U_n$ is naturally a geometric object, so it strikes me as unnatural to look for a geometric feel to it. In fact I sense that finite ring structures altogether are not amenable to satisfactory visualization in the sense of actual pictures. – anon May 12 '13 at 06:58

2 Answers2

4

In the case of the $n$-th roots of unity, consider the regular $n$-gon with vertex set $$ R = \{e^{\frac{2\pi ik}{n}}: k = 0, \dots, n-1\}. $$ Then $U_{n}$ is the set of the vertices in $R$ which are not vertices of any regular $k$-gon, for $1 \le k < n$ dividing $n$, whose vertex set $T$ satisfies $1 \in T \subset R$.

For instance when $n = 6$ the vertices are $$ R = \{ 1, e^{\frac{2\pi i}{6}}, e^{\frac{2\pi i}{3}}, -1, e^{\frac{- 2\pi i}{3}}, e^{\frac{- 2\pi i}{6}} \}. $$ For $k = 1$ we have the regular $1$-gon, with vertex $1$; $k = 2$ we have the regular $2$-gon of vertices $1, -1$; for $k = 3$ we have the equilateral triangle of vertices $1, e^{\frac{2\pi i}{3}}, e^{\frac{- 2\pi i}{3}}$.

So taking these vertices away, we are left with $$ U_6 = \{ e^{\frac{2\pi i}{6}}, e^{\frac{- 2\pi i}{6}} \}. $$

Andreas Caranti
  • 68,873
  • Thanks. A couple of things aren't clear to me. Firstly why is what you said a subgroup (there seems to be no identity, closure and inverse aren't clear either). Secondly, why is this the same as our traditional $U_n$? –  May 11 '13 at 11:22
  • @Shahab, mine is just an alternative description (with a geometric feel, I hope) for what you write as $U_n = {e^{\frac{2\pi ik}{n}}:(k,n)=1}$. I am simply removing the $e^{\frac{2\pi i k}{n}}$ with $(k, n) > 1$, so what's left is $U_{n}$. This is because if $(k, n) = t$, then $e^{\frac{2\pi ik}{n}} = e^{\frac{2\pi ik/t}{n/t}}$ lies on a regular $n/t$-gon inside the regular $n$-gon. – Andreas Caranti May 11 '13 at 13:10
  • I think I understand. But why are you removing $1$? –  May 12 '13 at 04:53
  • 1
    @Shahab: The root of unity $1$ corresponds to the zero element in ${\bf Z}/n{\bf Z}$, which is not invertible under multiplication. Keep in mind that the group operation on $U_n$ (which is multiplication in ${\bf Z}/n{\bf Z}$) is not at all obvious from this geometric description. In fact multiplication at all in the integers modulo $n$ is not obvious from the representation of the integers mod $n$ as the $n$th roots of unity in the complex plane. – anon May 12 '13 at 06:51
  • @anon, very good point. – Andreas Caranti May 12 '13 at 06:59
  • @anon: Isn't the binary operation ordinary multiplication in the case of the cyclic group of order n being understood as the nth roots of unity? So the identity element should be $1$. If we interpret $U_n$ as a subgroup of the nth roots of unity, $1$ should be within $U_n$. If the binary operation is not ordinary multiplication then what is it? –  May 12 '13 at 08:27
  • 1
    @Shahab: Please read what I wrote again. The root of unity $1$ corresponds to $0$ in the integers modulo $n$ (since $1=e^0$); these are the identity elements. The residue $1$ in the integers modulo $n$ (which is not an identity element) corresponds to $e^{2\pi i/n}$. In the roots of unity, the group operation is multiplication, and in the integers modulo $n$, the group operation is *addition*. Observe: $$\exp\left(2\pi i\frac{a}{n}\right)\color{Red}{\times}\exp\left(2\pi i\frac{b}{n}\right)=\exp\left(2\pi i\frac{c}{n}\right)\iff a\color{Red}{+}b\equiv c\bmod n$$ – anon May 12 '13 at 08:54
  • 1
    Now, the integers mod $n$ on top of having an addition operation also have their own multiplication operation. This means that ${\bf Z}/n{\bf Z}$ is something called a "ring." If you want to transport this operation over to the roots of unity, it would look like this: $$\exp\left(2\pi i\frac{a}{n}\right)\color{Red}{\bullet}\exp\left(2\pi i\frac{b}{n}\right)=\exp\left(2\pi i\frac{ab}{n}\right).$$ The roots of unity are not a group under this operation, though $e^{2\pi i/n}$ does function as an identity (just like how $1\in U_n$ is the identity) under $\bullet$. – anon May 12 '13 at 08:58
  • @anon: Got it. Thanks. –  May 12 '13 at 09:05
1

The chinese remainder theorem essentially says that $U_m$ is the product of $U_{p^n}$ for $p^n$ the exact power in which the prime $p$ divides $m$.

vonbrand
  • 27,812