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I have a problem linking the multiplication in a finite field with its additive structure:

The set $S$ is an additive subgroup of $\mathbb{F}_{2^h}$ (the finite field of order $2^h$). I have two different cosets, $\beta+S$ and $\gamma+S$. The question is, whether there is there always an element $\delta$, different from $0,1$ such that $$\delta(\beta+S)\cap (\gamma+S)=\emptyset.$$ In other words, can I multiply the coset with some element $\delta\neq 0,1$ and still be disjoint from the other coset?

In the case I was interested in (a particular additive subgroup of index $4$), I checked by computer for $h=4,5,6$ and there, it is always possible to find such a $\delta$. But I wonder whether there is a general reason for this, independent of the additive subgroup itself and the primitive polynomial.

Edit: the subgroup S I mentioned is simply the additive subgroup generated by $1,\alpha,\alpha^2$, where $\alpha$ is the primitive element $Z(2^h)$ that GAP uses. So I took $\beta=\alpha^3$ and $\gamma=\alpha^4$.

Choky
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    It is a little bit surprising that this would happen for large subgroups $S$. Think: two non-parallel planes in 3D-space, they always intersect non-trivially. An extreme example would be when $S$ has index two, and the cosets $\beta+S$,$\gamma+S$ cover the entire space seamlessly. Therefore I suspect that $S$ and $\delta S$ must not be in any kind of general position. It should be easier to achieve this, if $S$ were stable under multiplication by elements of an intermediate subfield $K$, $\Bbb{F}2\subset K\subset \Bbb{F}{2^h}$. Obviously such a field does not exist when $h=5$ (or any prime) – Jyrki Lahtonen Jul 13 '16 at 07:26
  • But, it may well be that I'm trying to unravel this from the wrong end. Is the description of the subgroup $S$ you are primarily interested in too long to include here? Meanwhile, I continue to try and think of a counterexample other than $[\Bbb{F}_{2^h}:S]=2$. – Jyrki Lahtonen Jul 13 '16 at 07:31

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