Yes.

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, 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?)

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.

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

let z+q be an element in Q/Z, while and ,

then , while and .

Then,

therefore, <<Why is that?! What is the neutral element in Q/Z? I don't understand this>>

If you have any coset you can reach the neutral element by adding it to itself b-times, because

.

Therefore every element in Q/Z has finite order.