How to compute the rational group of this elliptic curve? How to compute the rational group of this elliptic curve:
$$E:\quad y^2=(x+3)x(x-1).$$
Ps:  I am not familar with elliptic curves. (1,0), (0,0), (-3,0), (-1, 2), (-1, -2), (3, 6), (3, -6) are obviously rational points of E(Q), I guess that E(Q) is isomorphic to Z4×Z2, but I cannot prove it.

Yes, you are right. The rank is $0$, and the torsion subgroup is isomorphic to $E_{tor}=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, with generators $(3,6)$, $(0,1)$. 
Hence $E(\mathbb{Q})\simeq E_{tor}$.
Substituting $x$ by $x-1$ we obtain the minimal Weierstrass equation
$$
E\colon y^2=x^3-x^2-4x+4.
$$
All integer points are given by your above list. 
There are several results you can use to prove this, such as Nagell-Lutz for the torsion
subgroup, and for the rank the fact that the L-series value 
$$L(E,1)=0.539128911875$$ is nonzero (the order of vanishing is the analytic rank of $E$, which in this case is known to be the rank of $E$).
One can also check this by a computation in the online database (with $E=[0,2,0,-3,0]$) 
http://www.lmfdb.org/EllipticCurve/Q.
