In a field with $2^n$ elements what is the subgroup of squares?
I know that, for every prime power $p^n$, a field of order $p^n$ exists, but I don-t know how determine this subgrup. Can give me any hint, thanks!
Determine subgroup of squares.
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finite-fields
2 Answers
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the multiplicative group has order $2^{n}-1$ and is cyclic, the squares in a cyclic group of odd order are all of the elements of the group. This proves every invertible element is a square and clearly $0$ is also a square.
We conclude every element in the field is a square.
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If you don't know that the multiplicative group of a finite field is cyclic, then you can note that in any field $K$, $x^2 = y^2$ iff $(x - y)(x + y) = 0.$ In a field with characteristic $2$, $x + y = x -y$, so $x^2 = y^2$ iff $(x - y)^2 = 0$ iff $x = y$. So the map $x \mapsto x^2$ is one-to-one, for any $K$ of characteristic $2$, and hence onto, if $K$ is finite.