Abstract-Algebra-1-Cyclic Groups, guaranteed and verified 100% Pass

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Cyclic Groups
If 𝐺 = < 𝑎 >; i.e. 𝐺 = {𝑎𝑛 | 𝑛 ∈ ℤ} then 𝐺 is called a cyclic group and 𝑎 is its
generator. If 𝐺 is finite then the order of 𝑎 is the order of < 𝑎 >. If 𝐺 is not
finite we say 𝑎 has infinite order.


Ex. If 𝐺 = ℤ10 then 𝑎 = 1 generates 𝐺 and has order 10.
Ex. If 𝐺 = ℤ then 𝑎 = 1 generates 𝐺 and has infinite order.


Theorem: Every cyclic group is abelian.


Proof: 𝐺 = {𝑎𝑛 | 𝑛 ∈ ℤ}

Let 𝑔1 , 𝑔2 ∈ 𝐺 then 𝑔1 = 𝑎𝑘 , 𝑔2 = 𝑎 𝑗 .

𝑔1 𝑔2 = 𝑎𝑘 ∙ 𝑎 𝑗 = 𝑎𝑘+𝑗 = 𝑎 𝑗+𝑘 = 𝑎 𝑗 ∙ 𝑎𝑘 = 𝑔2 𝑔1 so 𝐺 is abelian.


Theorem (Division Algorithm for ℤ): If 𝑚 is a positive integer and 𝑝 is any integer,
then there exist unique integers 𝑞 and 𝑟 such that

𝑝 = 𝑚𝑞 + 𝑟 and 0 ≤ 𝑟 < 𝑚.

𝑝


−2𝑚 −𝑚 0 𝑚 2𝑚 3𝑚
If 𝑝 = 𝑞𝑚 for some 𝑞 then 𝑟 = 0. Otherwise 𝑞𝑚 < 𝑝 < (𝑞 + 1)𝑚 for some 𝑞
and since the distance between 𝑞𝑚 and (𝑞 + 1)𝑚 is 𝑚, 0 < 𝑟 < 𝑚.

We call 𝑞 the quotient and 𝑟 the non-negative remainder when 𝑝 is divided by 𝑚.

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Ex. Find 𝑞 and 𝑟 when 46 is divided by 7.


7, 14, 21, 28, 35, 42, 49, …
42 < 46 < 49
So 46 = 7(6) + 4

𝑞 = 6 and 𝑟 = 4.


Theorem: A subgroup 𝐻 of a cyclic group 𝐺 is cyclic.



Proof: Let 𝐺 be generated by 𝑎; 𝐺 = {𝑎𝑛 | 𝑛 ∈ ℤ}.

If 𝐻 = < 𝑒 > = {𝑒} then 𝐻 is cyclic.

If 𝐻 ≠ {𝑒} then 𝑎𝑝 ∈ 𝐻 for some 𝑝 ∈ ℤ+ .

Let 𝑚 be the smallest integer in ℤ+ such that 𝑎𝑚 ∈ 𝐻.



Now let’s show that 𝑑 = 𝑎𝑚 generates 𝐻.

We must show that every 𝑏 ∈ 𝐻 is a power of 𝑑.

Since 𝑏 ∈ 𝐻 and 𝐻 ≤ 𝐺, we have 𝑏 = 𝑎𝑝 for some 𝑝.

Find 𝑞, 𝑟 such that 𝑝 = 𝑚𝑞 + 𝑟 for 0 ≤ 𝑟 < 𝑚.

Then 𝑎𝑝 = 𝑎𝑚𝑞+𝑟 = (𝑎𝑚 )𝑞 𝑎𝑟 ,

so 𝑎𝑟 = (𝑎𝑚 )−𝑞 𝑎𝑝 .

Now since 𝑎𝑝 ∈ 𝐻, 𝑎𝑚 ∈ 𝐻 and 𝐻 is a group, both (𝑎𝑚 )−𝑞 and 𝑎𝑝
are in 𝐻 thus,

(𝑎𝑚 )−𝑞 𝑎𝑝 ∈ 𝐻 Thus 𝑎𝑟 ∈ 𝐻.