Prove Cyclic Factor Group of a Cyclic Group is Cyclic

[Fisicks]

Prove that the factor group of a cyclic group is cyclic.

Now I understand introductory AA very well. But I am just having a brain fart on this problem lol.

Let G= <g> and H be a subgroup of G. Then it is true, from basically definition that G/H= <gH>.
Where the equality means they are the same set but <gH> is going to repeat some cosets most of the time. Can i conclude G/H is cyclic?

If not then I would probably say take the duplicates out, and show that <gH> (without duplicates) is a subgroup of <gH> which i could then conclude G/H is cyclic.

 

[Bacle]

Hi, Fisicks:

don't mean to be pedantic, but , to make sure I understand you: you mean the
quotient group of a cyclic group G by _any_ subgroup is cyclic, right?

If so, as you said, G/H is defined to be the set {Hg: g in G} (you can select

any representative of Hg if Hg'=Hg; one can show G/H is well-defined in this

sense.) , and group operation given by: (Hg)(Hg')=Hgg' ; where gg' is the multiplication

in G.

Now, G is cyclic, so there is a g in G so that {g,g²,...,gⁿ,..

} =G (G may be cyclic and finite too; if so, the argument is similar )

Now, consider Hg , where g is the generator of G, and consider the product:

Hg_iHg_j:=Hg_ig_j

Can you see how to generate {Hg: g in G} this way?

 

[Bacle]

Just another proof, just in case:

i) Every infinite cyclic group is isomorphic to Z, the integers. Every finite cyclic group
is isomorphic to {nZ: n an integer}. Every quotient group of a cyclic group is then
of the form:

Z/nZ:~Z/n ={1,...,n-1,n} , which is generated by 1

Proof of above assumption on form of subgroups:

Every cyclic group is countably-infinite, i.e., can be indexed by the pos. integers.

But the pos. integers are well-ordered, so that every subgroup H--as a set-- has

a least element h. By group properties, if h is in H, then so is h+h=2h, ..., so is

h+h+...+h=nh,... . So every subgroup of G is of the form H={nh: h in H}.

An adjustment can be made for finite cyclic groups, using the fact that the

order of a subgroup divides the order of the group.


And, re the cosets, there is overlap of cosets in G/H, if |H|>1 , by Lagrange's

thm: |G/H|=|G|/|H|<|G| if |H|>1 , and |G/H|=|G| only iff |H|=1
All subgroups of a cyclic


My algebra is not the strongest, so better to listen to mathwonk

 

[mathwonk]

look at the image of the generator.

 

[Fisicks]

Mathwonk, you are saying that I cannot conclude that G/H is cyclic from the fact alone that G/H and <gH> have the same elements?

So if I prove that the image of <gH> is a subgroup of <gH>, I then know that G/H is cyclic because G/H is precisely the image of <gH> and any subgroup of a cyclic group is cyclic.

 

[Fisicks]

I just confused myself even more. I proved that G/H= H U gH U... U g^(k-1)H. Each is distinct because each power of g is less than k. So G/H=<gH>.

Someone clear my head up please. I know this is trivial so I want to move past this lol.

 

[Bacle]

I did not mean to imply i believed my answer was incorrect, only that mathwonk
is likely to be more insightful, while at my rudimentary level, I need to appeal
to the big machinery.

Anyway, a couple of comments:

1) Do a small quotient like S₃/A₃ , to see why
you don't get all different cosets.


2)G/H={Hg: g in G} is a group * , with multiplication defined by:

(Hg)(Hg'):=(Hgg')

If g is a generator for G, then, for any w in G, there is an integer n with

gⁿ =w

Now, can you see what you get by doing the product :

(Hg)(Hg)...(Hg) (n times)

What if you wanted to obtain the coset Hk , for some k in G. What would

you do; how would you use the generator g to do so?


* When H is normal in G, which is true in our case