In a set of lecture notes, there is an example of calculating the group $E_\text{tors}$ of an elliptic curve. This is the example:
Let $E$ be the elliptic curve $$y^2=x^3-5x+4.$$ The curve has discriminant $$\Delta=16(4\cdot 125 - 27\cdot 16)=2^6\cdot 17.$$ Therefore, if $[a:b:1]$ is a torsion point, we must have $$b=0,1,2,4,\text{or}\ 8.$$ Now the case $b=0$ clearly occurs with the point $[1:0:1]$. Let $f(x)$ denote the function $x^3-5x+4$. This function has derivative $3x^2-5$, which is positive for $|x|\geq 2$. Also, we have $$f(-3)=-8,\quad f(5)=104.$$ Combining these two statements we see the only possible integer values for $x$ for which $f(x)$ is between $0$ and $64$ are $$x=-2,-1,0,1,2,3,4.$$ Computing each of these two statements we see that the only possible torsion points are $$[0:\pm 2: 1],[3:\pm 4:1],[0:1:0],[1:0:1].$$ Also, one computes that $$[0:2:1]+[1:0:1]=[3:4:1].$$ However, if you compute $$[0:2:1]+[0:2:1]$$ then you find that the $x$-coordinate is $25/16$. As this is not an integer, by the Lutz-Nagell thereom $[0:2:1]$ is not a torsion point, and therefore $[3:4:1]$ is not a torsion point either. This implies that $E_\text{tors}=\{[0:1:0],[1:0:1]\}\simeq\mathbb{Z}/(2)$. This also show that the rank of $E$ is $\geq 1$.
So I understand why $[0:2:1]+[1:0:1]=[3:4:1]$, but it seems to me that there's the implicit assumption that $[0:1:0]$ and $[1:0:1]$ are already in the torsion group. Thus we know that since $[0:2:1]$ is not in the group, neither is $[3:4:1]$, for when composed with the inverse of $[1:0:1]$, it would give an element not in $E_\text{tors}$, which would mean $E_\text{tors}$ is not closed under composition. My question is, is it obvious that $[0:1:0]$ and $[1:0:1]$ are in the torsion group? Otherwise, I don't see why it would follow from $[0:2:1]+[1:0:1]=[3:4:1]$ that $[3:4:1]$ is not in torsion group, if we don't necessarily already know that $[1:0:1]$ is, and hence it's inverse, is in $E_\text{tors}$.
Also, is it generally true that if $[a:b:1]$ is not a torsion point, then neither is $[a,-b:1]$? This seems to be assumed as well. Thanks!