# Thread: Quotient Groups - Infinite Groups, finite orders

1. ## Quotient Groups - Infinite Groups, finite orders

I have a question: when I find a quotient group, I actually find the group of all cosets? I mean, every element in a quotient group is actually a coset?

For example, I have a confusing question:
Let (Q,+) be the rational group, (Z,+) be the integer a subgroup of Q.
Prove that every element in Q/Z has a finite order. (Does this mean I need to prove that every coset has a finite number of elements? or, what?)

Anyway, I googled this question, and I found a proof for that:

let z+q be an element in Q/Z, while $\displaystyle q \in Q$ and $\displaystyle z \in Z$,
then $\displaystyle q=\frac{a}{b}$, while $\displaystyle a,b \in Z$ and $\displaystyle b!=0$.
Then, $\displaystyle b*(z+q)=bz+qb=bz+a \in Z$

therefore, $\displaystyle |z+q|=b$ <<Why is that?! What is the neutral element in Q/Z? I don't understand this>>

Thank you very much

2. Originally Posted by adam63
I have a question: when I find a quotient group, I actually find the group of all cosets? I mean, every element in a quotient group is actually a coset?
Yes.

For example, I have a confusing question:
Let (Q,+) be the rational group, (Z,+) be the integer a subgroup of Q.
Prove that every element in Q/Z has a finite order. (Does this mean I need to prove that every coset has a finite number of elements? or, what?)
No, an element of a group is said to have "finite order", if you can reach the neutral element by multiplying( i.e. applying the group operation) the element to itself a finite number of times.
For example the element 1 in Z has NOT finite order, because you can never reach 0 just by adding 1 to itself.
In fact no element in Z except 0 itself has finite order.

Anyway, I googled this question, and I found a proof for that:

let z+q be an element in Q/Z, while $\displaystyle q \in Q$ and $\displaystyle z \in Z$,
then $\displaystyle q=\frac{a}{b}$, while $\displaystyle a,b \in Z$ and $\displaystyle b!=0$.
Then, $\displaystyle b*(z+q)=bz+qb=bz+a \in Z$

therefore, $\displaystyle |z+q|=b$ <<Why is that?! What is the neutral element in Q/Z? I don't understand this>>
The neutral element in Q/Z is the coset 0+Z=Z. In general the neutral element of any quotient group G/N is the coset 0+N=N.

If you have any coset $\displaystyle \frac{a}{b}+\mathbb{Z}, b>0$ you can reach the neutral element by adding it to itself b-times, because
$\displaystyle b\cdot\frac{a}{b}+\mathbb{Z}=a+\mathbb{Z}=0+\mathb b{Z}$.
Therefore every element in Q/Z has finite order.

3. Oh, beautiful, I understand it

Only one last question:

if I multiply b*(a/b+Z), shouldn't I get: a+bZ, which is a different coset?

*Now I understand - it's because you only add Z again and again... which is an infinite group, and you will always have the 'full' Z, no matter how many times you multiply it

4. Originally Posted by adam63
Oh, beautiful, I understand it

*Now I understand - it's because you only add Z again and again... which is an infinite group, and you will always have the 'full' Z, no matter how many times you multiply it
Exactly. The notation b*(a/b+Z) can be a bit confusing, so i edited out the paranthesis now. What is meant is indeed just adding the whole coset a/b+Z b-times to itself. You can also operate only on a representative a/b of the coset and then you get a representative b*a/b for the new coset. Note that a coset can have several representatives. Every element of the coset is a representative and two representatives x,y are equivalent, if x-y is in Z.

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# math forum is every group whose all elements are of finite order is finite

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