Understanding the Set of Fourth Roots of Unity

The set of fourth roots of unity consists of all complex numbers $z$ that satisfy the equation $z^4 = 1$. Solving this equation yields the roots $1, -1, i, -i$. So, the set of fourth roots of unity is $S = \{1, -1, i, -i\}$. We will analyze if this set forms a group under different operations.

Group Properties Under Addition for the Set of Fourth Roots of Unity

For a set to form a group under an operation, it must satisfy four conditions: closure, associativity, identity element, and inverse elements. Let's check the set of fourth roots of unity under the operation of addition.

Closure: If we add any two elements from the set $S$, the result must also be in $S$. Let's take two elements, say $1 \in S$ and $i \in S$. Their sum is $1 + i$. The complex number $1+i$ is not an element of the set $S = \{1, -1, i, -i\}$. Since the set is not closed under addition, the set of fourth roots of unity does not form a group with respect to sum.

Group Properties Under Multiplication for the Set of Fourth Roots of Unity

Now let's examine if the set of fourth roots of unity forms a group under multiplication.

• Closure: We need to ensure that the product of any two elements in $S$ is also in $S$. Let's create a multiplication table for the elements in the set of fourth roots of unity:

As shown in the table, all resulting products $\{1, -1, i, -i\}$ are elements of the set $S$. Therefore, the set of fourth roots of unity is closed under multiplication.

• Associativity: Multiplication of complex numbers is associative. For any $a, b, c$ in the set $S$, the property $(a \times b) \times c = a \times (b \times c)$ holds true.
• Identity Element: An identity element $e$ is an element in $S$ such that for any element $a \in S$, $a \times e = e \times a = a$. The element $1$ is in $S$, and for any element $a \in S$, $a \times 1 = 1 \times a = a$. Thus, $1$ is the multiplicative identity element.
• Inverse Element: For every element $a$ in $S$, there must exist an inverse element $a^{-1}$ in $S$ such that $a \times a^{-1} = a^{-1} \times a = 1$.
  • The inverse of $1$ is $1$, and $1 \in S$.
  • The inverse of $-1$ is $-1$, and $-1 \in S$ (since $(-1) \times (-1) = 1$).
  • The inverse of $i$ is $-i$, and $-i \in S$ (since $i \times (-i) = 1$).
  • The inverse of $-i$ is $i$, and $i \in S$ (since $(-i) \times i = 1$).
  Every element in the set has its inverse within the set.

Since the set of fourth roots of unity satisfies closure, associativity, has an identity element, and every element has an inverse under multiplication, it forms a group with respect to multiplication.

Is the Set of Fourth Roots of Unity an Abelian Group under Multiplication?

An abelian group is a group where the operation is commutative, meaning $a \times b = b \times a$ for all elements $a, b$ in the set. Multiplication of complex numbers is commutative. Looking at the multiplication table for the set of fourth roots of unity, it is symmetric about the main diagonal, confirming that the operation is commutative for these elements. Therefore, the group formed by the set of fourth roots of unity under multiplication is an abelian group.

Based on this analysis, the set of fourth roots of unity forms an abelian group with respect to multiplication, but not a group with respect to sum.