Number of $\alpha\in\Bbb{F}_{27}$ such that $|A_{\alpha}|=26$ equals $12$

Question

Question: Let $\Bbb{F}_{27}$ denote the finite field of size $27$. For each $\alpha\in\Bbb{F}_{27},$ we define $$A_{\alpha}=\{1,1+\alpha,1+\alpha+\alpha^2,1+\alpha+\alpha^2+\alpha^3,\dots\}.$$ Then show that number of $\alpha\in\Bbb{F}_{27}$ such that $|A_{\alpha}|=26$ equals $12$.

What I could conclude is,

$0\in A_{\alpha}$ if and only if $ \alpha\neq0$.

Then I thought that, $|A_{\alpha}|=26$ implies $A_{\alpha}=\Bbb{F}_{27}^*$, but I am not sure. Then some guess what I was making from the fact $\phi(26)=12$, but no progress. Help me with this, because I not getting any clue from the structure of $A_{\alpha}$.

Answer 1

$\DeclareMathOperator{ord}{ord}$ For any element $a\in\mathbb{F}_{27}^*$ denote the order of element $a$ in the multiplicative group $a\in\mathbb{F}_{27}^*$ as $\ord a$. We will prove that for all $\alpha\neq 1$ we have $|A_{\alpha}|=\ord\alpha$. Indeed, note that $$ |A_{\alpha}|=|(1-\alpha)\cdot A_{\alpha}|=|\{(1-\alpha)\cdot(1+\alpha+\ldots+\alpha^k)\mid k\geq 0\}|= \\ =|\{1-\alpha^{k+1}\mid k\geq 0\}|=|\{\alpha^m\mid m\geq 1\}|=\ord\alpha. $$

Therefore, you just need to find the number of elements of $a\in\mathbb{F}_{27}^*$ of order 26, which is exactly $\varphi(26)=12$, as you have mentioned in the question.

Note 1. For $\alpha=1$ we have $|A_1|=3$.

Note 2. Here, $c\cdot A$ means $c\cdot A=\{c\cdot a\mid a\in A\}$.