You are (slightly) misreading the statements. I've checked through all 13 occurrences of the word "order" in that paper, and see only "typical" statements, such as:
Let
be a group of prime order
"Order" means one of two things in group theory:
- The order of a group is just the number of elements in it. So a group of prime order
is just a group 
where 
. - The order of an element of a group
is how many times you have to repeat the group operation of 
to get to the identity. For a group written additively, it's the smallest 
such that 
. For a group written multiplicatively, its the smallest 
such that 
.
These are (sort of) connected by a result in group theory that says that 
. As a consequence, the order of an element always divides the order of the group.
This is a generalization of both Fermat's little theorem and Euler's theorem.
The paper you're reading only seems to use order to mean the first of these two things. Note that groups of prime order are special in that they're always cyclic groups.
The most basic example of a cyclic group is 
, and in fact every cyclic group is isomorphic to a group of this form. Note that this isomorphism is generally not efficiently computable, which is good because the discrete logarithm problem is easy in 
.
Videos
I ordered some headphones on Amazon, I could've sworn they were Prime but they won't be arriving today as expected and they immediately offered the 1-month free or $10 credit. Fine. But I wanted to know where I could see this information in my order details, emails, nor reports. You can see it at the time of purchase with the giant Prime symbol or if it says "fullfilled by amazon".
Asked the support lady how she knew, or confirmed, and she guided me through the Order history whether it says 'Sold by: Company vs Amazon.' - which I didn't believe because I had multiple orders, with screenshots at the time of purchase they didn't matchup to what she was telling me. (I know bullshit when it's right in front of my face)
It came down to me asking to speak with a supervisor or "guru". Basically he said the only way to tell was to know at the time of purchase if the order is Prime or not ('Sold by So&So Fullfilled by Amazon' = Prime). They say they can view this from their side, but no way for us to tell. I ran a report of all my purchases this year and knew which ones were prime or not. So happens that I had 2 more that were late, and she ended up giving me another $20 on top of today's realization. (I document my dates and take screenshots of everything!)
I asked for their team to create a helpdesk ticket or something to pass onto their team in which helps users view if previous orders were actually prime orders or not. "Of course" - he told me. He mentioned he would but then couldn't CC me or send me a ticket number stating a request has been documented. So, the odds are nothing will happen, and the suggestion wouldn't be communicated or documented. And we came to the realization that users have to call and waste both of our's time just to see this one line of info. (I can't confirm this of course since I don't work for Amz but like I said, I can smell bullshit.)
So basically I was fed a line from their frontline support person. And everything she told me didn't add up. Supervisor was a little more helpful but seems he could save his support team some time by sending a simple request for consideration to the higher ups or to management.
The way I see it, Amazon doesn't want to show you if an order is Prime, in which if your package is late, they'd have to cough up $10 or 1 month free membership. It'll cost the money.
If anyone knows another way to confirm if an previous order was prime (fullfilled by amazon) or not, let me know. Just thought I'd share.
Let $G$ be a group of prime order $p > 1$. Let $a \in G$ such that $a \neq e$. Note that $\langle a \rangle \leq G$, so by Lagrange's theorem $|\langle a \rangle|$ divides $|G|$. Since $p$ is prime, either $|\langle a \rangle| = 1$ or $|\langle a \rangle| = p$, but $|\langle a \rangle| = 1$ is not possible since that would imply $\langle a \rangle = \{e\}$ and therefore $a = e$. So $|G| = p = |\langle a \rangle|$, and therefore $G = \langle a \rangle$.
As Cam McLeman comments, Lagranges theorem is considerably simpler for groups of prime order than for general groups: it states that the group (of prime order) has no non-trivial proper subgroups.
I'll use the following
Lemma
Let $G$ be a group, $x\in G$, $a,b\in \mathbb Z$ and $a\perp b$. If $x^a = x^b = 1$, then $x=1$.
Proof: by Bezout's lemma, some $k,\ell\in\mathbb Z$ exist, such that $ak+b\ell=1$. Then $$ x = x^{ak+b\ell} = (x^a)^k \cdot (x^b)^\ell = 1^k \cdot 1^\ell = 1 $$
(If you know a little ring theory, you might prefer to notice that the set $\{i | x^i=1\}\subseteq \mathbb Z$ forms an ideal which must contain $(a,b)=1$ if it contains $a$ and $b$.)
The question
Now let $P$ be an arbitrary group of prime order $p$. Consider any $x\in P$ such that $x\neq 1$ and consider the set $$ S = \{ 1, x, x^2 , \dots , x^{p-1} \}\subseteq P.$$ First assume two of these elements are equal, say $x^u=x^v$ and $u<v$ without loss of generality. Then $x^{v-u}=1$ and $1\leq v-u \leq p-1$. But then surely $v-u \perp p$. By the lemma, $x^{v-u} = x^p = 1$ now implies that $x=1$, a contradiction so every two members of $S$ must be different.
But then $|S|=p$. This implies $S=P$ and $P=\langle x\rangle$ is cyclic.