Mazur's theorem-abelian torsion group of rational points of an elliptic curve I am looking at Mazur's theorem... 
$$E(\mathbb{Q})_{\text{torsion}} \cong \mathbb{Z}/n\mathbb{Z}, \text{ for } n=1,2, \dots ,10,12$$
means that the torsion group $E(\mathbb{Q})_{\text{torsion}}$ contains the points $P$ which have the identity
$$nP=0, \text{ for } n=1,2, \dots ,10,12$$
and the coordinates are integers, right ?? 
What does
$$E(\mathbb{Q})_{\text{torsion}} \cong \mathbb{Z}/2m\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}, \text{ for } m=1,2,3,4$$
mean?? 
 A: The result specifies what are the exact possibilities for the torsion part of the group of rational points on an elliptic curve (over the rationals). 
By Mordell's theorem you know that  $E(\mathbb{Q})_{\text{torsion}}$ is some finite abelian group. Now, you know that it cannot by just whatever finite abelian group but it must be one of a very particular form. 
Namely, the group is either cyclic with order $\le 10$ or equal to $12$ or it is a direct product of a group of order two with a cyclic group of order $2,4,6,$ or $8$. 
For one particular elliptic curve it can be only one of these groups of course, but the result asserts that for every elliptic curves you will get one of these groups. 
If the torsion part is cyclic, yes, this means about what you said: there is a point $P$ with rational coefficients on the group that has order $n$ where $n\le 10$ or $n=12$, and $P, 2P, \dots, nP=0$ is the set of all $n$ rational points of finite order on this curve.
For the other case, you have two points $P,Q$ the order of $P$ is $2$ the order of $Q$ is $n$ with $n$ equal $2,4,6,$ or $8$ and $Q,2Q, \dots ,n Q$ and $P+Q , P + 2Q, \dots , P+nQ$ is the set of all $2n$ rational points of finite order on this curve.
