If H is nonzero, c is a torsion element. Thus a torsion element is a bit like a zero divisor.
The torsion submodule MT is the submodule consisting of all torsion elements in M. We need to prove this is a submodule. If c is killed by x then yc is also killed by x, so scaling is no problem. If x kills c and y kills d, then xy kills c+d. Since R is an integral domain, xy is nonzero, and c+d is torsion. This makes MT a submodule.
A module is torsion free if it has no torsion elements, and torsion if it equals its torsion submodule.
Let h be a module homomorphism on M. If xc = 0 then xh(c) = h(xc) = 0, and the image of a torsion element c is torsion. If h(c) is torsion then h(xc) = 0, and xc lies in the kernel of h.
As a special case, consider M/MT. If h(c) is torsion then c maps into the kernel, which is torsion, hence c is torsion. Therefore the image is torsion free.
Let R be a pid and let p be a prime element, such that pi kills c. Let x generate the ideal that annihilates c. Thus x divides pi, and since R is a ufd, x = pj for some j ≤ i.
The generator of the annihilator of c is called the order of c. It is zero iff c is torsion free. The order of c cannot be a unit, unless c = 0, which isn't very interesting.
This process can be reversed. If x is an element of R, let M be the quotient R mod x. This is an R module that is killed by x. The module, as a whole, has order x.