If you don't specify more about the structure of the field $K$, then we can't say much about the structure of the group $E(K)$.  There are special cases (described in [the Wikipedia article][1]):

 1. If $K$ is a number field, then the [Mordell-Weil theorem][2] implies the group is finitely generated (and this has been generalized, as Anweshi mentioned).  In fact, for each number field, there is a global bound on the size of the torsion of any elliptic curve over that field.  In particular, if $K = \mathbf{Q}$, then it is a direct sum of a free abelian group of finite rank with a torsion group that is one of 15 types.
 2. If $K$ is finite of order $q$, then (by a theorem of Hasse) the group is finite of order about $q+1$ with error bounded by $2\sqrt{q}$.  It is a sum of two cyclic groups.

If $K$ is larger than that, then $E(K)$ can be quite large.  For example, if $K$ is separably closed, then $E(K)$ is divisible.  In this case, if $K$ has characteristic zero, then $E(K) \cong (\mathbf{Q}/\mathbf{Z})^2 \oplus \bigoplus \mathbf{Q}$.  If $K$ has characteristic $p>0$, then $E(K) \cong \bigoplus_{\ell \neq p} (\mathbf{Q}_\ell/\mathbf{Z}_\ell)^2 \oplus (\mathbf{Q}_p/\mathbf{Z}_p)^h \oplus \bigoplus \mathbf{Q}$.  Here, $h$ is zero or one depending on whether the curve is supersingular or ordinary, and the $\bigoplus \mathbf{Q}$ is a vector space whose dimension is:

 1. zero if K is an algebraic closure of a finite field.
 2. countably infinite if $K$ is countable and not an algebraic closure of a finite field.
 3. equal to the cardinality of $K$ otherwise.

Away from the separably closed case, you get a subgroup of one of these groups, but you can have very complicated subgroups of $\mathbf{Q}$ as summands, and very complicated torsion subgroups.

[1]: http://en.wikipedia.org/wiki/Elliptic_curve
[2]: http://en.wikipedia.org/wiki/Mordell-Weil_theorem