Is there a way to project one elliptic-curve element to a subgroup with certain size? For discrete logarithm we can pick a random number $n$ and project it to a subgroup. E.g. given a prime $p$ with
$p-1 = 2\cdot a \cdot b +1$
we can compute
$n^{((p-1)/a)} \equiv n_a \mod p$
after this $n_a$ is an element of a subgroup with size $a$
$(n_a)^a \equiv 1 \mod p$  

Is there a way to do something like this for elliptic curves?



Target use case scenario:
1.)random number $r$ is picked
2.)element of elliptic curve $e$ out of $r$ with some function gets computed: $e_r=f(r)$
3.)element which lies in subgroup with size $k$ gets computed: -> $e_{r,k}$  (how?)
4.)repeat 1.-3. for another random number $r$ --> $e'_{r',k}$
target result: A possible attacker who has access to source code and runtime variables should not know how to compute $e'_{r',k}$ out of $e_{r,k}$ with given EC generator $g$.

The subgroup with size $k$ likely (if exists) also has a generator $g_k$. There should be some way to produce $e'_{r',k}$ out of $e_{r,k}$, e.g. with $e'_{r',k} = e_{r,k} \odot g_k^i$. In case multiple subgroups with size $k$ exist which have disjoint elements this only need to be true for elements in same group. The amount of disjoint groups should not be larger than a low number (&lt;1000).

In case that is possible whould it have a significant impact at security?
Is there a way to do something like this for elliptic curves?


Of course; if we have an elliptic curve with order $q = a \cdot b$, then we can make an element $H$ into the subgroup of order $a$ by computing $H_a = (q/a)H$.  And, we have $a H_a = 0$ (the neutral element) precisely as requested.

Now, we typically don't use elliptic curves with an order with two large prime factors; they are usually selected with an order $q = hr$, where $h$ is a small integer (most often 1, 4 or 8) and $r$ is a large prime.  We do this because that makes the ECDLP problem as hard as possible; with your order $q = a \cdot b$, you could compute discrete logs in time $\sqrt{ \max( a, b ) }$, and so we'd have to use a larger elliptic curve that we would otherwise need.