Properties:
- {e} = <e>
- ℤ+ = {0, ±1, ±2, ±3, ... } = <1>
- ℚ+, ℝ+, are not cyclic.
- S2 = {(1) , (12)} = <(12)> ≠ <(1)>
- Sm , m > 2, is not cyclic.
- The multiplicative group {1, w, w2} formed by the cube roots of unity is a cyclic group.
- The multiplicative group {1, -1, i, -i } formed by the fourth roots of unity is a cyclic group.
- The group of integers under addition is an infinite cyclic group generated by 1.
Example: The multiplicative group {1, w, w2} formed by the cube roots of unity is a cyclic group.
Sol. Let G be the group of cube roots of unity under multiplication.
∴ G = {1, w, w2}.
Here, 1 = w3, therefore each element of G is an integral power of w.
∴ G is cyclic group generated by w.
i.e., G = <w>.
Example: The multiplicative group {1, -1, i, -i } formed by the fourth roots of unity is a cyclic group.
Sol. G = {1, -1, i, -i } is a group under multiplication.
Since 1 = (i)4, -1= i2, -i = i3.
Therefore each element of G is an integral power of i.
∴ G is a cyclic group generated by i
i.e., G = <i>.
Example: The group of integers under addition is an infinite cyclic group generated by 1.
Sol. Since each n ∈ ℤ can be written as
n = n · 1.
∴ ℤ = <1>.
Also each n ∈ ℤ can be written as
n = (-n)(-1) where -n ∈ ℤ.
∴ ℤ = <-1>
i.e., ℤ has two generators 1 and -1.
Example: The group G = {(0, 1, 2, 3, 4, 5), +6} is a cyclic group under the operation addition congruence modulo 6.
Sol. Since 1 = 1(1)
2 = 1 +6 1 = 2(1)
3 = 1 +6 1 +6 1 = 3(1)
4 = 1 +6 1 +6 1 +6 1 = 4(1)
5 = 1 +6 1 +6 1 +6 1 +6 1 = 5(1)
0 = 1 +6 1 +6 1 +6 1 +6 1 +6 1 = 6(1)
∴ Every element n of G can be written as
n = n(1).
∴ G = <1>
i.e., G is a cyclic group.
Note: Here G is also generated by 5, i.e., G = <5>.
∴ A group may have more than one generators.
Example: Give an example of abelian group which is not cyclic.
Sol. Let <ℚ , +> be the group of rational numbers under addition.
It is an abelian group which is not cyclic.
For, suppose that (m/n) ∈ ℚ is a genrator of ℚ.
Then every element of ℚ should be an integral multiple of m/n.
Let (1/3n) ∈ ℚ be any element.
Let (1/3n) = k · (m/n) for some integer k.
⇒ (1/3) = km.
Which is not possible as k, m are integers, where as (1/3) is not.
Hence no element can act as generator of ℚ.
Example: The multiplicative group {1, -1, i, -i } formed by the fourth roots of unity is a cyclic group.
Sol. G = {1, -1, i, -i } is a group under multiplication.
Since 1 = (i)4, -1= i2, -i = i3.
Therefore each element of G is an integral power of i.
∴ G is a cyclic group generated by i
i.e., G = <i>.
Example: The group of integers under addition is an infinite cyclic group generated by 1.
Sol. Since each n ∈ ℤ can be written as
n = n · 1.
∴ ℤ = <1>.
Also each n ∈ ℤ can be written as
n = (-n)(-1) where -n ∈ ℤ.
∴ ℤ = <-1>
i.e., ℤ has two generators 1 and -1.
Example: The group G = {(0, 1, 2, 3, 4, 5), +6} is a cyclic group under the operation addition congruence modulo 6.
Sol. Since 1 = 1(1)
2 = 1 +6 1 = 2(1)
3 = 1 +6 1 +6 1 = 3(1)
4 = 1 +6 1 +6 1 +6 1 = 4(1)
5 = 1 +6 1 +6 1 +6 1 +6 1 = 5(1)
0 = 1 +6 1 +6 1 +6 1 +6 1 +6 1 = 6(1)
∴ Every element n of G can be written as
n = n(1).
∴ G = <1>
i.e., G is a cyclic group.
Note: Here G is also generated by 5, i.e., G = <5>.
∴ A group may have more than one generators.
Example: Give an example of abelian group which is not cyclic.
Sol. Let <ℚ , +> be the group of rational numbers under addition.
It is an abelian group which is not cyclic.
For, suppose that (m/n) ∈ ℚ is a genrator of ℚ.
Then every element of ℚ should be an integral multiple of m/n.
Let (1/3n) ∈ ℚ be any element.
Let (1/3n) = k · (m/n) for some integer k.
⇒ (1/3) = km.
Which is not possible as k, m are integers, where as (1/3) is not.
Hence no element can act as generator of ℚ.
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