Let $\mathbb F_{27}$ be the finite field of order $27$. Let $A_{a}=\{1,1+a,1+a+a^2,\dots\}$. Then which of the following are true?   The number of $a\in\mathbb F_{27}$ such that $\operatorname{card}(A_{a})=26$ equals to $12$. $0 \in A_{a} \iff a$ is nonzero. $\operatorname{card}(A_1)=27$ $\bigcap A_{a}$ is a singleton.   For 2 one way is very obvious. If $0=a$ then $0=1+a+a^2+... +a^k$ for some $k &lt;= 27$.Put a=0 and we get $1=0$, which is a contradiction. For the other way round I am unable to proceed.  For 4 I have no ideas how to proceed.
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  the number of  $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.  $ 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
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