# Let $(G,*)$ be a group with $a \in G$ such that $\operatorname{Ord}(a)=p$ with $p$ being prime numbers.

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 $$o(a^k) ≠ p \forall 1\leq k. Then, $$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 $$k with $$p$$ prime, then $$\gcd(k,p)=1$$, so $$\operatorname{Ord}(a^k)=p \forall 1\leq k.

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 $$m. Thus, $$\gcd(m,p)=1$$. Hence, $$o(a^m) = \frac{p}{1}= p$$.

Thanks for correction!

• 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".. Commented Mar 1, 2020 at 15:06
• sorry ........? Commented Mar 1, 2020 at 15:08

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 $$a^m=e$$.