The elliptic curve group comes with formulae for adding together points with different formulae for distinct points and doubling. In your example, we can hand calculate the formulae by hand. We begin with $(2,3)+(2,3)$ so we use the doubling formula. We calculate the slope
$$s=\frac{3x_P^2+a}{2y_P}=\frac{3\times 2^2+0}{2\times 3}=2$$
and use this to compute new $x$ and $y$ coordinates
$$x_R=s^2-2x_P=2^2-2\times 2=0;\quad\quad y_R=y_P+s(x_R-x_P)=3+2\times(0-2)=-1.$$
We change the sign on $y_R$ to complete the group law. Thus $2(2,3)=(0,1).$ Next we compute $(0,1)+(2,3)$ using the distinct points formula. Again, we start with the slope
$$s=\frac{y_P-y_Q}{x_P-x_Q}=\frac{1-3}{0-2}=1$$
and use this to compute new $x$ and $y$ coordinates
$$x_R=s^2-x_P-x_Q=1^2-0-2=-1;\quad\quad y_R=y_P+s(x_R-x_P)=1+1\times(-1-0)=0$$
Changing the sign on $y_R$ doesn't change it so that $3(2,3)=(-1,0)$.
Similarly, $(-1,0)+(2,3)=(0,-1)$ so that $4(2,3)=(0,-1)$ and $(0,-1)+(2,3)=(2,-3)$ so that $5(2,3)=(2,-3)$.
Lastly we note that our formulae do not return rational numbers in the case $x_P=x_Q$ and $y_P=-y_Q$. In this special case, we say that the points add up to the point at infinity which is the group identity. Thus $6(2,3)=\mathcal O$ and the element is of order6 and so the subgropu that it generates is also of order 6.
