Manipulating additive subgroups of the finite field GF($2^m$) Consider the field construction as in this link, where GF($2^m$) is obtained as a concatenation of additive subgroups of cardinality $2^j$ for $j=0,1,...,m$. In this question, no assumption is made that subgroups of cardinality $2^{j'}$ for $j'$ that divides $s$ need to be subfields.
Assume that two subgroups $\mathcal{S}_a$ and $\mathcal{S}_{b}$ are chosen (following the notations in the link above, where $\mathcal{S}_a$ is of cardinality $2^a$ and $\mathcal{S}_b$ is of cardinality $2^b$), where w.l.o.g. $\mathcal{S}_a \subseteq \mathcal{S}_b$ and $a+b \le m$. Can we always find an element $g \ne 0$ of the field such that $\{g \cdot {S_a}\}\bigcap {{S_b}}  = \left\{ 0 \right\}$?
Note that $g \cdot {S_a}$ is an additive subgroup of GF($q$). Considering the subgroups as subspaces over GF($2$), my first thought was to look for $g$, such that when multiplied by the basis elements of $\mathcal{S}_a$, lead to a disjoint set of basis elements. However, I am not sure how to do this in a systematic way.

Addendum:
For fixed values of $a,b$ such that $a+b \le m$ such $g$ can indeed be found. However, is there a "universal" value of $g$ such that
$$\{ g \cdot {S_a}\} \bigcap {{S_b}}  = \left\{ 0 \right\}$$
$$\{ {S_a}\} \bigcap {\left\{ {g \cdot {S_b}} \right\}}  = \left\{ 0 \right\}$$
for every pair $\mathcal{S}_a,\mathcal{S}_b$ such that $a+b \le m$?
 A: It is enough to consider $1\leq a$ and  $a+b=m$, otherwise it becomes trivial case. Let $\{v_1, \ldots , v_a\}$ be a basis for $S_a$, and $\{w_1, \ldots, w_b\}$ be a basis for $S_b$. Let $F^*$ be the set of nonzero elements in $GF(2^m)$. 
We find an element $g\in F^*$ satisfying:
$$
g \cdot v_1 \notin S_b.\ \ \ (1)
$$
Since multiplication by nonzero element is injective, there are at least $2^m - 1-2^b$  elements $g$ satisfy the above. 
We also want $g$ to satisfy:
$$
g\cdot v_2 \notin \mathrm{Span}(S_b \cup \{g\cdot v_1\}). \ \ \ (2)
$$
The negation of this condition is in fact the existence of $a_{b+1}\in \{0,1\}$ such that 
$$
g\cdot v_2 = a_1 w_1 + \cdots + a_b w_b + a_{b+1} g\cdot v_1
$$
which is equivalent to 
$$
g \cdot (v_2 - a_{b+1} v_1) \in S_b.
$$
We avoid all $g$ possibly satisfying the above. There are two choices for $a_{b+1}$ and $v_2 - a_{b+1} v_1 \neq 0$. Thus, multiplication by $v_2 - a_{b+1} v_1$ is injective. Thus, there are at least $2^m-1-2^b-2^{b+1}$ elements $g$ satisfy both (1) and (2). 
We use this idea  to find $g\in F^*$ satisfying (1) and:
For all $1\leq j\leq a-1$, 
$$
g\cdot v_{j+1} \notin \mathrm{Span}(S_b \cup \{g\cdot v_1 , \ldots , g\cdot v_j\}). \ \ \ (j+1)
$$
Then the negation of the above is satisfied by at most $2^{j+b}$ elements $g\in F^*$. Therefore there are at least $2^m-1-\sum_{j=b}^{m-1} 2^j$ elements $g\in F^*$ satisfying (1) and ($j+1$) for all $1\leq j\leq a-1$. 
We have $2^m-1-\sum_{j=b}^{m-1} 2^j\geq 1$. Thus, finding such $g$ is possible. 
