Consider the set $G=\left\{ {0,1,...,{2^m} - 1} \right\}$. The elements of this set can be viewed as the elements of GF($q=2^m$) with appropriate addition/multiplication operations. For example, GF(4) = $\left\{ {0,1,2,3} \right\}$, GF(8)=$\left\{ {0,1,2,3,4,5,6,7} \right\}$ with field operations defined in GF(4), GF(8) (where $a,b,...$ should be understood here as $2,3,...$.). Note that $2^m$ above should be understood as "regular" exponentiation (i.e., the exponent operation in $\mathcal{R}$). In particular, $G$ can be viewed as an Abelian group with the addition operation of the field.
It is clear that subgroups of $G$ must be of sizes that are a power of $2$, according to Lagrange's theorem. Define the sets ${S_j} = \left\{ {0,1,...,{2^j} - 1} \right\}$ for $j=1,2,...,m$ (again, $2^j$ is a regular exponentiation, e.g. ${S_2} = \left\{ {0,1,2,3} \right\}$). I have the following questions:
- Can the addition operation of the field GF($q=2^m$) can always be defined such that $S_j$ are subgroups of $G$ (it is the case in the examples above)?
- If so, $g \cdot {S_j} = \left\{ {g \cdot 0,g \cdot 1,...,g \cdot \left( {{2^j} - 1} \right)} \right\}$ for $g \in G$ is also a subgroup of $G$. is it correct to say that all subgroups of $G$ are of the form $g \cdot S_j$?
- Otherwise, which subgroups exist except for $g \cdot S_j$?
Here is my comment expanded as an answer.
First, to define the addition to make all the $S_j$ subgroups, we can do the following: Write the numbers from $0$ to $2^m-1$ in binary and define the addition bitwise (mod $2$). Thus, we see that $S_j$ consists of those numbers that have $0$'s on the first (i.e. most significant) $m-j$ bits, and clearly this subset is closed under the chosen addition.
To see that not all subgroups are of the form $S_j$ when $m\geq 4$, we can simply count the number of subgroups of order $4$. A subgroup of order $4$ is determined by picking two distinct elements, but there are $3$ choices of such two elements for each subgroup. So the number of subgroups of order $4$ is $\frac{1}{3}\binom{2^m-1}{2} = \frac{(2^m-1)(2^m-2)}{6} > 2^m-1$ (since $2^m-2 > 6$). But there are only $2^m-1$ subgroups of the form $g\cdot S_2$, so these cannot be all the subgroups of order $4$ (and the other $S_j$ have different orders).