What is the order of each element of the group?

The Mathematician — Sun, 13 Nov 2022 13:00:31 +0000

In this article, we will explain what the order of each element of the group.

Definition 1. Let

G

be a group. The order of an element

x \in G

is the smallest positive integer

n

such that

x^{n} = e

, where

e

is the identity element.

We can show several examples depending on which order the group

G

is defined. We take here for example the additive group

(\mathbb{Z}/10\mathbb{Z}, +)

, the multiplicative group

((\mathbb{Z}/10\mathbb{Z})^{\times}, \times)

and the symmetric group

S_3

.

Example 1. Take the additive group

(\mathbb{Z}/10\mathbb{Z}, +)

. Here the identity element is

e = 0

. Further, we need to check for which

x \in \mathbb{Z}/10\mathbb{Z}

we get that

xn = 0

for

n \in \mathbb{Z}/10\mathbb{Z}

. So we get for each element the next order:

For 0, we get $\{0 \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{0\}$ . So we have that the order of 0 is 1.
For 1, we get $\{n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{1,2,3,4,5,6,7,8,9,0\}$ . So the order of 1 is 10.
For 2, we get $\{2n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{2,4,6,8,0\}$ . So the order of 2 is 5.
For 3, we get $\{3n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{3,6,9,2,5,8,1,4,7,0\}$ . So the order of 3 is 10.
For 4, we get $\{4n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{4,8,2,6,0\}$ . So the order of 4 is 5.
For 5, we get $\{5n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{5,0\}$ . So the order of 5 is 2.
For 6, we get $\{6n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{6,2,8,4,0\}$ . So the order of 6 is 5.
For 7, we get $\{7n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{7,4,1,8,5,2,9,6,3,0\}$ . So the order of 7 is 10.
For 8, we get $\{8n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{8,6,4,2,0\}$ . So the order of 8 is 5.
For 9, we get $\{9n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{9,8,7,6,5,4,3,2,1,0\}$ . So the order of 9 is 10.

Example 2. Take the group

((\mathbb{Z}/10\mathbb{Z})^{\times}, \times)

. We know that

(\mathbb{Z}/10\mathbb{Z})^{\times} = \{1,3,7,9\}

. So we get

For 1, we get $\{1^n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{1\}$ . So the order of 1 is 1.
For 3, we get $\{3^n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{3^1,3^2,3^3,3^4\} = \{3,9,7,1\}$ . So the order of 3 is 4.
For 7, we get $\{7^n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{7^1,7^2,7^3,7^4\} = \{7,9,3,1\}$ . So the order of 7 is 4.
For 9, we get $\{9^n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{9^1,9^2\} = \{9,1\}$ . So the order of 9 is 2.

Example 3. Our last example is the symmetric group

S_3

which is closed under composition. Recall that

\begin{equation*} S_3 = \{e,(12),(13),(23),(123),(132)\}. \end{equation*}

Obviously,

e

has order 1. The cycles of length two has order 2, e.g.,

(12)^2 = (12)(12) = e

. The cycles of length three has order 3, e.g.,

(123)^3 = (123)(123)(123) = (123)(132) = e

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