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∈G is the smallest positive integer
n such that
xn=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
(Z/10Z,+), the multiplicative group
((Z/10Z)×,×) and the symmetric group
S3.
Example 1. Take the additive group
(Z/10Z,+). Here the identity element is
e=0. Further, we need to check for which
x∈Z/10Z we get that
xn=0 for
n∈Z/10Z. So we get for each element the next order:
- For 0, we get {0 ∣ n∈Z/10Z}={0}. So we have that the order of 0 is 1.
- For 1, we get {n ∣ n∈Z/10Z}={1,2,3,4,5,6,7,8,9,0}. So the order of 1 is 10.
- For 2, we get {2n ∣ n∈Z/10Z}={2,4,6,8,0}. So the order of 2 is 5.
- For 3, we get {3n ∣ n∈Z/10Z}={3,6,9,2,5,8,1,4,7,0}. So the order of 3 is 10.
- For 4, we get {4n ∣ n∈Z/10Z}={4,8,2,6,0}. So the order of 4 is 5.
- For 5, we get {5n ∣ n∈Z/10Z}={5,0}. So the order of 5 is 2.
- For 6, we get {6n ∣ n∈Z/10Z}={6,2,8,4,0}. So the order of 6 is 5.
- For 7, we get {7n ∣ n∈Z/10Z}={7,4,1,8,5,2,9,6,3,0}. So the order of 7 is 10.
- For 8, we get {8n ∣ n∈Z/10Z}={8,6,4,2,0}. So the order of 8 is 5.
- For 9, we get {9n ∣ n∈Z/10Z}={9,8,7,6,5,4,3,2,1,0}. So the order of 9 is 10.
Example 2. Take the group
((Z/10Z)×,×). We know that
(Z/10Z)×={1,3,7,9}. So we get
- For 1, we get {1n ∣ n∈Z/10Z}={1}. So the order of 1 is 1.
- For 3, we get {3n ∣ n∈Z/10Z}={31,32,33,34}={3,9,7,1}. So the order of 3 is 4.
- For 7, we get {7n ∣ n∈Z/10Z}={71,72,73,74}={7,9,3,1}. So the order of 7 is 4.
- For 9, we get {9n ∣ n∈Z/10Z}={91,92}={9,1}. So the order of 9 is 2.
Example 3. Our last example is the symmetric group
S3 which is closed under composition. Recall that
S3={e,(12),(13),(23),(123),(132)}.
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.