Proof of Finite Order of G in Quotient Group Q/Z

[Bellarosa]

1. Show that every element of the quotient group G = Q/Z has finite
order. Does G have finite order?
he problem statement, all variables and given/known data



2. This is the proof

The cosets that make up Q/Z have the form Z + q,
where q belongs to Q. For example, there is a coset Z + 1/2, which is
the set of all numbers of the form {n + 1/2}, where n is an integer.
And there is a coset Z + 2/5, which consists of all numbers of the
form n + 2/5, where n is an integer. The cosets form a group if you
define the sum of A and B to be the set of all sums of an element in A
and an element in B, and this group is the quotient group. The
identity of Q/Z is just Z.

Now if you take a rational number r/s, where r and s are integers,
then

s (r/s) = r

which is an integer. Now anything in the coset Z + r/s is an integer
plus r/s, so if you multiply anything in that coset by s, you get an
integer. So if you multiply the coset by s (i.e. add it to itself s
times) you get a coset consisting of all integers, but that's just Z
itself. That is, the coset is of finite order s (or a divisor of s).



3.I just need it to be explained especially the part with s(r/s)

 

[morphism]

What you said is perfectly correct. To put it in precise terms, if r/s+Z is an element of Q/Z (where we may assume that s is a positive integer), then s(r/s+Z)=r+Z=Z. So every element of Q/Z has finite order.

 

[Bellarosa]

ok... I can see by example why this proof make sense with s in r/s being a prime number because s here is the order of the factor group, how about when s is not prime, does the proof still apply?

 

[morphism]

Why would s have to be a prime number? The order of what factor group? Q/Z is definitely not of order s, or any other finite number!

 

[Bellarosa]

ok...