by Bethany Nylander

Definition: The elements of the set *U _{6}* = {z $\in$ $\mathbb{C}$ $|$ z

^{6}= 1} are called the

**6**.

^{th}roots of unityAccording to the definition of **complex numbers**, $\mathbb{C}$, we recall that z represents a + bi where a,b $\in \mathbb{R}$.

By letting *U _{6}* be the circle in the Euclidean plane with center at the origin and radius 1, we see that the elements of the set

*U*are the numbers

_{6}cos $\left(m\frac{2\pi}{6}\right)$ +

*i*sin $\left(m\frac{2\pi}{6}\right)$ for m = 0, 1, 2, 3, 4, 5.

Namely, the elements of this set are:

- cos 0 +
*i*sin 0

- cos $\left(\frac{\pi}{3}\right)$ +
*i*sin $\left(\frac{\pi}{3}\right)$

- cos $\left(\frac{2\pi}{3}\right)$ +
*i*sin $\left(\frac{2\pi}{3}\right)$

- cos $\left(\pi\right)$ +
*i*sin $\left(\pi\right)$

- cos $\left(\frac{4\pi}{3}\right)$ +
*i*sin $\left(\frac{4\pi}{3}\right)$

- cos $\left(\frac{5\pi}{3}\right)$ +
*i*sin $\left(\frac{5\pi}{3}\right)$

which all have absolute value 1. If we let $\zeta$ = cos $\left(\frac{2\pi}{6}\right)$ + *i* sin $\left(\frac{2\pi}{6}\right)$, then these 6^{th} roots of unity can be written as

1 = $$\zeta$^{0}, $$\zeta$^{1}, $$\zeta$^{2}, $$\zeta$^{3}, $$\zeta$^{4}, $$\zeta$^{5} complex numbers respectively.

The image below represents these roots of unity on the unit circle.

$$\zeta$^{0} is represented as z_{0}, $$\zeta$^{1} is represented as z_{1}, $$\zeta$^{2} is represented as z_{2}, $$\zeta$^{3} is represented as z_{3}, $$\zeta$^{4} is represented as z_{4}, and $$\zeta$^{5} is represented as z_{5}.

Geometrically, we multiply complex numbers, such as $$\zeta$^{0}, $$\zeta$^{1}, $$\zeta$^{2}, $$\zeta$^{3}, $$\zeta$^{4}, $$\zeta$^{5}, by multiplying their absolute values and adding their polar angles. The absolute values of the 6^{th} roots of unity are all 1, and we know 1^{n} = 1 for all n $\in \mathbb{Z}$. The polar angle between any 6^{th} roots of unity is n$\left(\frac{\pi}{3}\right)$ for some n $\in \mathbb{Z}$ with addition modulo 2$\pi$, as can be seen in the above unit circle diagram. Since the multiplication of the absolute values of the 6^{th} roots of unity are closed under multiplication, and the addition of the polar angles of the 6^{th} roots of unity are closed under addition modulo 2$\pi$, then **the elements in the set U_{6} are closed under complex multiplication**.

Also, because $\zeta$^{6} = 1, we see these 6 powers of $\zeta$ are closed under multiplication. For example, we have

$$\zeta$^{4}$$\zeta$^{5} = $$\zeta$^{9} = $$\zeta$^{6}$$\zeta$^{3} = 1 $\times$ $$\zeta$^{3} = $$\zeta$^{3}.

We will now turn to prove that the binary algebraic structure of set *U _{6}* with complex multiplication is in fact a group.

We will prove that the binary algebraic structure of set *U _{6}* with complex multiplication, $\star$, is a group (

*U*, $\star$).

_{6}We have shown in the Introduction to the Group section that the set *U _{6}* is closed under complex multiplication $\star$.

- Let a, b, c $\in \mathbb{Z}$ be arbitrary, and therefore let $$\zeta$
^{a}, $$\zeta$^{b}, $$\zeta$^{c}$\in$*U*be arbitrary. Then by the property of complex multiplication of exponentials, we have_{6}

- $$\zeta$
^{a}$\star$ ($$\zeta$^{b}$\star$ $$\zeta$^{c}) = $$\zeta$^{a}$\star$ ($$\zeta$^{b+c}) - = $$\zeta$
^{a+(b+c)} - By the property of associativity in integer addition, we have $$\zeta$
^{a+(b+c)}= $$\zeta$^{(a+b)+c} - = ($$\zeta$
^{a+b}) $\star$ $$\zeta$^{c} - ($$\zeta$
^{a}$\star$ $$\zeta$^{b}) $\star$ $$\zeta$^{c}- Because for all $$\zeta$
^{a}, $$\zeta$^{b}, $$\zeta$^{c}$\in$*U*where a, b, c $\in \mathbb{Z}$ are arbitrary, we have $$\zeta$_{6}^{a}$\star$ ($$\zeta$^{b}$\star$ $$\zeta$^{c}) = ($$\zeta$^{a}$\star$ $$\zeta$^{b}) $\star$ $$\zeta$^{c}, associativity holds, meeting the first group axiom.

- Because for all $$\zeta$

- Also from the Introduction to the Group section, we know that an element in the set
*U*is (cos 0 +_{6}*i*sin 0) = 1 = $$\zeta$^{0}. As in the sets $\mathbb{R}$ and $\mathbb{Z}$ with multiplication, the identity element for $\mathbb{C}$ with multiplication is the number 1. Since the elements in*U*are complex numbers and the operation is complex multiplication, the identity element for this group is 1 also, or to be more notationally correct the identity element in_{6}*U*is $$\zeta$_{6}^{0}. For all a $\in \mathbb{Z}$ is arbitrary and therefore all $$\zeta$^{a}$\in$*U*be arbitrary, we have_{6}

- $$\zeta$
^{0}$\star$ $$\zeta$^{a}= $$\zeta$^{0+a}= $$\zeta$^{a} - AND $$\zeta$
^{a}$\star$ $$\zeta$^{0}= $$\zeta$^{a+0}= $$\zeta$^{a}, showing that the identity element is $$\zeta$^{0}for $\star$.

- We show that each element in
*U*, namely ($$\zeta$_{6}^{0}, $$\zeta$^{1}, $$\zeta$^{2}, $$\zeta$^{3}, $$\zeta$^{4}, and $$\zeta$^{5}) has an inverse in*U*equaling the identity element $\zeta$_{6}^{0}.

- For element $\zeta$
^{0}in*U*, there is the inverse element $\zeta$_{6}^{0}in*U*such that $\zeta$_{6}^{0}$\star$ $$\zeta$^{0}= $\zeta$^{0+0}= $\zeta$^{0}, the identity element. - For element $\zeta$
^{1}in*U*, there is the inverse element $\zeta$_{6}^{5}in*U*such that $\zeta$_{6}^{1}$\star$ $\zeta$^{5}= $\zeta$^{1+5}= $\zeta$^{6}= 1 = $\zeta$^{0}, the identity element. - For element $\zeta$
^{2}in*U*, there is the inverse element $\zeta$_{6}^{4}in*U*such that $\zeta$_{6}^{2}$\star$ $\zeta$^{4}= $\zeta$^{2+4}= $\zeta$^{6}= 1 = $\zeta$^{0}, the identity element. - For element $\zeta$
^{3}in*U*, there is the inverse element $\zeta$_{6}^{3}in*U*such that $\zeta$_{6}^{3}$\star$ $\zeta$^{3}= $\zeta$^{3+3}= $\zeta$^{6}= 1 = $\zeta$^{0}, the identity element. - For element $\zeta$
^{4}in*U*, there is the inverse element $\zeta$_{6}^{2}in*U*such that $\zeta$_{6}^{4}$\star$ $\zeta$^{2}= $\zeta$^{4+2}= $\zeta$^{6}= 1 = $\zeta$^{0}, the identity element. - For element $\zeta$
^{5}in*U*, there is the inverse element $\zeta$_{6}^{1}in*U*such that $\zeta$_{6}^{5}$\star$ $\zeta$^{1}= $\zeta$^{5+1}= $\zeta$^{6}= 1 = $\zeta$^{0}, the identity element.

Because all three group axioms are satisfied, we have shown that (*U _{6}*, $\star$) is indeed a group.

Subgroups of the sixth roots of unity with complex number multiplication, (*U _{6}*, $\star$):

- the 6th roots of unity, {z $\in$ $\mathbb{C}$ $|$ z
^{6}= 1}, with complex number multiplication - (*U*, $\star$) - the improper subgroup_{6} - the 3rd roots of unity, {z $\in$ $\mathbb{C}$ $|$ z
^{3}= 1}, with complex number multiplication - (*U*, $\star$) - a proper and nontrivial subgroup_{3} - the 1st root of unity, {z $\in$ $\mathbb{C}$ $|$ z
^{1}= 1}, with complex number multiplication - (*U*, $\star$) - the trivial subgroup_{1}

It is obvious by the definition of a subgroup that (*U _{6}*, $\star$) $\leq$ (

*U*, $\star$) as

_{6}*U*$\subseteq$

_{6}*U*and (

_{6}*U*, $\star$) is a group itself.

_{6}Similarly, *U _{3}* $\subset$

*U*, as can be seen by comparing the graph below with the graph in the Introduction to the Group section, and (

_{6}*U*, $\star$) is a group itself.

_{3}Therefore by the definition of a subgroup, (

*U*, $\star$) $\leq$ (

_{3}*U*, $\star$).

_{6}And lastly, *U _{1}* $\subset$

*U*, and (

_{6}*U*, $\star$) is a group itself.

_{1}Therefore by the definition of a subgroup, (

*U*, $\star$) $\leq$ (

_{1}*U*, $\star$).

_{6}I think (*U _{6}*, $\star$) is isomorphic to ($\mathbb{Z}$

_{6}, +

_{6}).