Prove that the number of $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|=26 $ equals 12.

Let $\mathbb{F}_{27}$ denote the finite field of size 27. For each $\alpha\in\mathbb{F}_{27}$ , define $ A_{\alpha}$ $=$ { $ 1, 1+\alpha , 1+\alpha+{\alpha}^2 , 1+\alpha+{\alpha}^2+{\alpha}^3 , .... $} . Then prove that

1. the number of $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.

2. $ 0\in A_\alpha $ if and only if $\alpha\neq 0$

   For the first proof , the multiplicative group of the field $\mathbb{F}_{27}$ is cyclic and its order is 26 , so there are 12 generators in that group.

Now I need to show that $ 1+\alpha+{\alpha}^2+.. {\alpha}^{25} $ =0 , where order of $\alpha$=26 .

Since $\alpha ^{26}$ =1

($\alpha$-1)(1+$\alpha$+${\alpha}^2$+...+${\alpha}^{25}$)=0 ,

since $\alpha\neq 1$ , So $ 1+\alpha+...+{\alpha}^{25} $ =0

My question is , can I write these last two lines? I mean are those two lines alright to write?

If those two lines are invalid then I need some help to show $ 1+\alpha+{\alpha}^2+..+{\alpha}^{25} $=0 .Thanks

Solution 1:

$\varphi (26)=12$ so there are $12$ generators, and indeed the $1+\alpha +\dots+\alpha ^k=\dfrac{1-\alpha ^{k+1}}{1-\alpha}\,,0\le k\le 25$ are all distinct for any of those generators.

(That $1+\alpha+\dots +\alpha ^{25}=0$ is clear since the sum equals $\dfrac{1-\alpha^{26}}{1-\alpha}$ and $\alpha ^{26}=1$.)