# Math Help - 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 $q \in Q$ and $z \in Z$,
then $q=\frac{a}{b}$, while $a,b \in Z$ and $b!=0$.
Then, $b*(z+q)=bz+qb=bz+a \in Z$

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

Thank you very much

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

therefore, $|z+q|=b$ <<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 $\frac{a}{b}+\mathbb{Z}, b>0$ you can reach the neutral element by adding it to itself b-times, because
$b\cdot\frac{a}{b}+\mathbb{Z}=a+\mathbb{Z}=0+\mathb b{Z}$.
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