I have a doubt conscerning the validity of my reasonning mostly the part "-2" .
Question:
We say that en element $g$ of a group $G$ is of torsion if it exists $n>0$ such that $g^n=e$ ($e$ the identity element of $G$). Show that all the element of $ \mathbb{Q} / \mathbb{Z} $ are of torsion.
My answer:
1-
We have the group $(\mathbb{Q};+)$. It is trivial to prove that $\mathbb{Z}$ is a normal sub group of $(\mathbb{Q};+)$ so $ ( \mathbb{Q} / \mathbb{Z} ; + ) $ is a group.
In the group $ \mathbb{Q} / \mathbb{Z} $ two rationnal numbers $q_1, q_2$ are considered equivalent iff $ q_1-q_2 \in \mathbb{Z} $. Hence we can writte that the set $ \mathbb{Q} \cap [0;1) $ is "sufficient for describing" $\mathbb{Q} / \mathbb{Z}$ as it includes in a simple way all the left coset that are possible (and this is the definition of $\mathbb{Q} / \mathbb{Z}$).
2-
Hence let focus on a specific $ q' \in \mathbb{Q} \cap [0;1) \Rightarrow q'= \frac{a}{b}, b>a \in \mathbb{N} $
It is obvious that here $q'^n=q'+...+q'=nq'$ and so if we take $n = b \in \mathbb{N} $ we get that this $q'$ is of torsion as $bq'=b \frac{a}{b}=a \in \mathbb{N} $
But $a \in \mathbb{Q} / \mathbb{Z}$ so it is equivalent to $0 \in \mathbb{Q} \cap [0;1)$ which is theidentity element.
3-
This reasonning is correct for any $q'$ so Q.E.D.
Is it correct?
As I said at the beginning I have a doubt mostly on the second part ("-2") but of course if you see error before I will be happy to know it so I can improve myself.
Thank you.
The idea is correct. Though there is no real reason to reduce ourselves to representatives in $[0,1)$. Also, better to work with elements of the group (i.e cosets), not by describing equivalence classes. A general element has the form $\frac{a}{b}+\mathbb{Z}$ where $a\in\mathbb{Z}, b\in\mathbb{N}$. Then:
$b\cdot(\frac{a}{b}+\mathbb{Z}) =\frac{ba}{b}+\mathbb{Z}=a+\mathbb{Z}=\mathbb{Z}.$
Which means $\frac{a}{b}+\mathbb{Z}$ is a torsion element.