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Let $(G,*)$ be a group with $a \in G$ and $\operatorname{Ord}(a)=p$ where $p$ is prime. (a) Prove that $\operatorname{Ord}(a^k) = p \forall 1\leq k < p$. (b) Prove that $\forall m \in \mathbb{N}$, either $a^m = e$ or $\operatorname{Ord}(a^m) = p$.

Please correct about my explanation below. (a) i used the contradiction.. Let $o(a^k) ≠ p \forall 1\leq k<p$. Then, $a^{pk} \neq e$. Thus, $(a^p)^k \neq e \Rightarrow e^k \neq e$. $\Leftrightarrow e * e * e* \cdots * e$ which recurse in $k$ times. Hence, $e ≠ e$. Contradiction. Therefore, $\operatorname{Ord}(a^k) = p$.

secondly, i used this theorem. Theorem. Let $(G,*)$ be a group with $a \in G$ such that $\operatorname{Ord}(a)=n. \forall t \in ℕ, \operatorname{Ord}(a^t) = \frac{n}{\gcd(t,n)}$.

From this theorem, $\forall k \in \mathbb{N}, o(a^k) = \frac{p}{\gcd(k,p)}$. Since $k<p$ with $p$ prime, then $\gcd(k,p)=1$, so $\operatorname{Ord}(a^k)=p \forall 1\leq k<p$.

for (b), i used the definition.. By definition, $\forall m \in \mathbb{N},$ $\operatorname{Ord}(a)=m \Rightarrow (a^m) = e$.

By theorem, since $\operatorname{Ord}(a)=p$, then $\forall, m \in \mathbb{N}, o(a^m)=\frac{p}{\gcd(m,p)}$ with all prime number $p$ and $m<p$. Thus, $\gcd(m,p)=1$. Hence, $o(a^m) = \frac{p}{1}= p$.

Thanks for correction!

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  • $\begingroup$ the negation of $\forall k $ between $1$ and $p $ is that there exists an element $a^k$ with $Ord(a^k) \ne p$ and not as you wrote it "for all".. $\endgroup$
    – infinity
    Commented Mar 1, 2020 at 15:06
  • $\begingroup$ sorry ........? $\endgroup$
    – lap lapan
    Commented Mar 1, 2020 at 15:08

1 Answer 1

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For the case b, note that for any positive integer $m$, $$\gcd(m,p)=\begin{cases} p & \text{if }p|m \\ 1&\text{if }p\not| m \end{cases}$$ therefore for all $m$, $$\text{ord}(a^m)=\frac{p}{\gcd(m,p)}=\begin{cases} 1 & \text{if }p|m \\ p &\text{if }p\not| m \end{cases}$$ therefore either $\text{ord}(a^m)=p$ or $\text{ord}(a^m)=1$ which implies $a^m=e$.

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  • $\begingroup$ what about case (a) ? is this correct? $\endgroup$
    – lap lapan
    Commented Mar 2, 2020 at 1:58
  • $\begingroup$ @arnold The case (a) is correct. $\endgroup$
    – Qurultay
    Commented Mar 2, 2020 at 8:38

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