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in group theory, an elementary abelian group is a finite abelian group, where every nontrivial element has order $p$, where $p$ is a prime; it is a particular kind of $p$-group. now suppose that we have a finite field $GF(p^n)$ and we want to consider it as a vector space over $GF(p)$. Can we claim that $GF(p^n)$ is an elementary abelian group? I have already read the answer of "Amitesh Datta". Finite fields as vector spaces

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Nil
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An elementary abelian group $\;G\;$ is one for which $\;pg=0\;$ for some prime $\;p\;$ and for all $\;g\in G\;$, i.e. any non-trivial element has (additive, in my writing) order equal to $\;p\;$

Now ask yourself whether it is true that $\;pa=0\;\;\forall\,a\in GF(p^n)\;$ ...

DonAntonio
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  • so for elements of GF(4) which are 0,1,2,3, the order of 3 is not equal to 2. right? – Nil Feb 06 '14 at 12:02
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    Those are not the elements of $;GF(4);$ , @user118746 ...!I'm afraid you're confussing the field with 4 elements and the non-field ring $;\Bbb Z/4\Bbb Z=:\Bbb Z_4;$ with four elements. – DonAntonio Feb 06 '14 at 12:08
  • GF(p) is called the prime field of order p , and is the field of residue classes modulo , where the elements are denoted 0, 1, ...,p-1 . ? The finite field GF(2) consists of elements 0 and 1? – Nil Feb 06 '14 at 12:20
  • Yes to both your questions. Only when $;n=1;$ we have that equality $;GF(p)=\Bbb Z/p\Bbb Z;$, @user118746 – DonAntonio Feb 06 '14 at 12:28
  • @user118746: Look up this question for the arithmetic of $GF(4)$. Some aspects of the arithmetic operations in $GF(8)$ and $GF(16)$ are spelled out here. – Jyrki Lahtonen Feb 06 '14 at 12:33