Question about kernel of reduction mapping of elliptic curves 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 + $gcd(m,p) = 1) \implies p \nmid b$ ? )

There can be points of infinite order in $E(\mathbb Q)$ whose coordinates are integral.  Also, as Alvaro Lozano-Robledo indicates, reduction mod $p$ need not
send $p$-torsion points to the identity.
