Let $E$ be an elliptic curve over $\mathbb{Q}$ which has good (nonsingular) reduction $\tilde{E}$ modulo some prime $p$. Denote their groups $E(\mathbb{Q})$ and $\tilde{E}(\mathbb{F}_p)$ respectively.
Basically the reduction is a map $r_p: E(\mathbb{Q}) \to \tilde{E}(\mathbb{F}_p)$ and because the reduction is good, this map is a group homomorphism. For a point $P$ it just reduces modulo $p$ the P's coordinates, if they are in $\mathbb{Z}$.
In case some coordinate of $P$ is in $\mathbb{Q}$ and has the form $\frac{a}{b}$, it can be properly reduced only if $b$ has multiplicative inverse modulo $p$ which is only in case when $p \nmid b$. Otherwise such points map to group's identity element. It follows that points with rational coordinates with modulo-$p$-non-invertible denominators belong to the mapping's kernel.
There's a theorem which states, that (for good reduction modulo $p$) the mapping is injective, when restricted to $m$-torsion points for all such $m$ that $gcd(m,p) = 1$.
The theorem seems to "dump" the kernel by assuming $gcd(m,p) = 1$ and thus ensuring the injectivity. This assumption seems natural, as elements having $m$-torsion with $m$ divisible by $p$ should intuitively all map to the group's identity element. It also dumps all torsion-free elements.
My question: Does this assumption "dump" only all points having a rational coordinate(s) with non-invertible denominators modulo $p$ (the kernel according to mapping definition)? Or does it dump a little bit more than that?
(In other words, are these assumptions equivalent: $(m$-torsion + $gcd(m,p) = 1) \iff p \nmid b$
or is the left one stronger: $(m$-torsion + $gcd(m,p) = 1) \implies p \nmid b$ ? )
Consider the following example. Let $E: y^2+y=x^3-x^2$. Its discriminant is $-11$, and $E(\mathbb{Q})\cong \mathbb{Z}/5\mathbb{Z}$, generated by $(0,0)$. The torsion points are, in fact: $$P=(0,0),\ 2P=(1,-1),\ 3P=(1,0),\ 4P=(0,-1),\ 5P=\infty=[0,1,0].$$ In particular, $p=5$ is a prime of good reduction, and the reduction mod $5$ of each of the points above is distinct. Moreover, $E(\mathbb{F}_5)$ consists precisely of the reduction of these $5$ points.
The point is that, in this case, $E(\mathbb{Q})[5]$ is not only defined over $\mathbb{Z}$ (so there are no $5$'s in the denominators), but moreover it injects into $E(\mathbb{F}_5)$. So $E(\mathbb{Q})[5]$ is not contained in the kernel of reduction modulo $5$.