What is the order of each element of the group?

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

Definition 1. Let GG be a group. The order of an element xGx \in G is the smallest positive integer nn such that xn=ex^{n} = e, where ee is the identity element.

We can show several examples depending on which order the group GG is defined. We take here for example the additive group (Z/10Z,+)(\mathbb{Z}/10\mathbb{Z}, +), the multiplicative group ((Z/10Z)×,×)((\mathbb{Z}/10\mathbb{Z})^{\times}, \times) and the symmetric group S3S_3.

Example 1. Take the additive group (Z/10Z,+)(\mathbb{Z}/10\mathbb{Z}, +). Here the identity element is e=0e = 0. Further, we need to check for which xZ/10Zx \in \mathbb{Z}/10\mathbb{Z} we get that xn=0xn = 0 for nZ/10Zn \in \mathbb{Z}/10\mathbb{Z}. So we get for each element the next order:
  • For 0, we get {0  nZ/10Z}={0}\{0 \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{0\}. So we have that the order of 0 is 1.
  • For 1, we get {n  nZ/10Z}={1,2,3,4,5,6,7,8,9,0}\{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  nZ/10Z}={2,4,6,8,0}\{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  nZ/10Z}={3,6,9,2,5,8,1,4,7,0}\{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  nZ/10Z}={4,8,2,6,0}\{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  nZ/10Z}={5,0}\{5n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{5,0\}. So the order of 5 is 2.
  • For 6, we get {6n  nZ/10Z}={6,2,8,4,0}\{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  nZ/10Z}={7,4,1,8,5,2,9,6,3,0}\{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  nZ/10Z}={8,6,4,2,0}\{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  nZ/10Z}={9,8,7,6,5,4,3,2,1,0}\{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 ((Z/10Z)×,×)((\mathbb{Z}/10\mathbb{Z})^{\times}, \times). We know that (Z/10Z)×={1,3,7,9}(\mathbb{Z}/10\mathbb{Z})^{\times} = \{1,3,7,9\}. So we get
  • For 1, we get {1n  nZ/10Z}={1}\{1^n \ | \ n \in \mathbb{Z}/10\mathbb{Z}\} = \{1\}. So the order of 1 is 1.
  • For 3, we get {3n  nZ/10Z}={31,32,33,34}={3,9,7,1}\{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 {7n  nZ/10Z}={71,72,73,74}={7,9,3,1}\{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 {9n  nZ/10Z}={91,92}={9,1}\{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 S3S_3 which is closed under composition. Recall that
S3={e,(12),(13),(23),(123),(132)}.\begin{equation*} S_3 = \{e,(12),(13),(23),(123),(132)\}. \end{equation*}
Obviously, ee has order 1. The cycles of length two has order 2, e.g., (12)2=(12)(12)=e(12)^2 = (12)(12) = e. The cycles of length three has order 3, e.g., (123)3=(123)(123)(123)=(123)(132)=e(123)^3 = (123)(123)(123) = (123)(132) = e.

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