-3
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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?

  1. The number of $a\in\mathbb F_{27}$ such that $\operatorname{card}(A_{a})=26$ equals to $12$.

  2. $0 \in A_{a} \iff a$ is nonzero.

  3. $\operatorname{card}(A_1)=27$

  4. $\bigcap A_{a}$ is a singleton.

For 2 one way is very obvious. If $0=a$ then $0=1+a+a^2+... +a^k$ for some $k <= 27$.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.

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marked as duplicate by Alexander Gruber abstract-algebra Aug 24 at 20:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ From my intuition I can say that 4 is true although I don't have any firm proof. $\endgroup$ – user481975 Aug 14 at 18:47
  • $\begingroup$ I did one too.Except the additive and multiplicative identity A_a will produce 26 elements and therefore 3 would be false too.If you kindly can provide the proof for 2 and 4 $\endgroup$ – user481975 Aug 14 at 18:57
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    $\begingroup$ Please include your thoughts and efforts on the problem. $\endgroup$ – Servaes Aug 14 at 19:11
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    $\begingroup$ Either this is a duplicate of this recent question or the extra cases may make this too broad, or it is lacking context :-/ $\endgroup$ – Jyrki Lahtonen Aug 14 at 19:35
  • $\begingroup$ Note that if $a$ is contained in a subfield, then the entire set $A_a$ is contained in that subfield, which shows that its cardinality is $\le 3$ for all three $a\in \mathbb F_3$. Note further that $A_0 =\lbrace 1\rbrace$, and $1$ is contained in each $A_a$, which already proves no. 4. $\endgroup$ – Torsten Schoeneberg Aug 14 at 19:37
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If $a=0$, then $A_0=\{1\}$ and if $a=1$, then $A_{1}=\{0,1,2\}$. For the following let us assume that $a \neq 0,1$.

An element of the form $1+a+a^2+\dotsb+a^{k-1}=(a^k-1)(a-1)^{-1}$ (note: $a-1$ is invertible in $\Bbb{F}_{27}$, because $a \neq 1$). Let us call the element $(a-1)^{-1}=b$. Since $a \in \Bbb{F}_{27}-\{0,1\}$, so $a^{26}=1$. Observe that we can write $$A_a=\{1,b(a^2-1),b(a^3-1), \ldots, b(a^{25}-1),0\}.$$ For $|A_a|=26$, we want all the elements in this list to be distinct.

Ques: When will $b(a^k-1)=b(a^j-1)$?

For this to occur, $a^{k-j} \equiv 1$. This means the order of $a$ is less than or equal to $k-j$ and $k-j<26$. Thus if order of $a$ is exactly $26$ then we will not have these collisions in the set $A_a$. Same as saying that $a$ must be a primitive root (or generator) of $\Bbb{F}_{27}^{\times}$.

So the number of such $a'$s is $\color{red}{\phi(26)=\phi(2)\phi(13)=12.}$

Part(2): If $a \neq 0$, then the order of $a$ is some positive integer $k$. In which case the element $b(a^k-1)=0$, thus $0 \in A_{a}$ for $a \neq 0$. I have already shown above that $0 \not\in A_0$.

Part(3): is false as shown above.

Part(4): is trivially true because $A_0=\{1\}$ has only one element.

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  • $\begingroup$ I can actually...Just need a little help upon the 4 $\endgroup$ – user481975 Aug 14 at 19:43
  • $\begingroup$ I recommend that you move your answer to the duplicate. It looks like this thread is on its way out. It would be already closed as a dupe if I had not accidentally picked another close reason (but this Laphroaig is fine) $\endgroup$ – Jyrki Lahtonen Aug 14 at 19:43
  • $\begingroup$ @JyrkiLahtonen Even though this has more parts than the duplicate question (I didn't notice the duplicate until I saw your note) but the basic theme remains the same. So if this gets closed so be it. I was just trying to help OP to better understand the idea. Moving this to duplicate doesn't seem to add anything extra there and the answers to other parts may add to the confusion. $\endgroup$ – Anurag A Aug 14 at 19:49

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