Any group of prime order is cyclic - Proof blueprint [Fraleigh p. 100 Cory 10.11]

Question

Not querying the proof or formality. I include only part of the proof.

The order of the group is a prime number. Call it p. Hence by means of the definition of prime number, $p > 1$. Since the order is prime, the group has > 1 element. It therefore has an element that is not the identity element. Call that element g. Then the group generated by g is $\langle g\rangle$. This is cyclic and has more than 1 element: g itself and the identity. Now $\langle g\rangle$ is a subgroup of G, and by Lagrange's Theorem, $| \langle g\rangle|$ divides $|G|$.

Why should one consider the subgroup generated by $g \neq id$, which is defined as $\langle g\rangle$? Is there an intuition as to why the proof uses this?

Let $G$ be a group and let $a \in G.$ Then $H = \{a^n : n \in \mathbb{Z} \}$ is a subgroup of G and is the smallest subgroup of G that contains a.

Answer 1

The fact that the group has prime order greatly restricts the orders of its subgroups, by Lagrange's theorem. Subgroups can only have order $1$ or order $p$. You want to get your hands on some nontrivial subgroup and study it. So you do the simplest possible thing: pick an element and look at the subgroup it generates.

I suppose the general philosophy is that you can understand a group by understanding its subgroups.

Answer 2

I think perhaps you're getting lost in the formality. The idea is this:

1. We are given a group $G$ such that $|G| = p$ for some prime $p$.

2. To show that $G$ cyclic, it suffices to show that there exists an element of order $p$ (why?).

3. So let's choose an arbitrary nonidentity element of the group, say $a$. What is the order of this element? Well, let's take powers of $a$ and see how many we need to wrap around to the identity.

4. Well, there are limited choices. The trick is recognizing that taking powers of $a$ forms a subgroup of $G$, which is denoted $\langle a\rangle$. We can now apply Lagrange's Theorem: $|\langle a\rangle|$ divides $|G|$. Since $|G|$ is prime, it's only nontrivial divisor is $p$.

We conclude $\langle a\rangle$ has $p$ distinct elements, which must be $\{e, a, a^2, a^3, ..., a^{p-1}\}$. With a bit of deductive reasoning, we can get $a*a^{p-1} = a^p = e$. This would imply that the element $a$ has order $p$, and thus $G$ is cyclic.

---

To directly answer your question, I can't really explain the "motivation" behind why I chose to look at the subgroup generated by $a$. There's simply a certain degree of intuition that you gain after studying a subject for a long enough time.

Answer 3

It might not be the best answer you could receive for you question- "I don't understand how you'd know to consider to prove this?". But, I suspect if you are asking something like one of the following common questions everybody asks, at least at the beginning, then I may have something to say.

• "How did we know a subgroup generated by only one element (different from the identity element) would be the right choice?"

• "Why did not we consider the subgroups generated by more than one element?"

• "How can the subgroup generated by this element "g" complete the proof? while the only thing we know about "g" is that g is not the identity element!!!"

Since you are working with a group of order p (a prime number), you want to combine your knowledge about the structure of a group and properties of prime numbers. (The last two answers have wisely pointed out this aspect).

But, we might still have something to think about. "What is the assumption without which the statement was wrong?" Or possibly "Could we replace this assumption with a weaker one?"

We want to show that the group of order p is cyclic. So, if, for a moment, we assume that we construct a subgroup generated by two elements (none of them identity), could we hope for a complete proof, without talking the relation between those two generators? The answer is no. Because, whatever you prove, is based on a subgroup generated by two elements, while a cyclic group is supposed to be generated by only one element.

There is still a question left. "Why did not we start with the identity element?" Let's assume we could complete the proof by the subgroup generated by the identity. It would mean that we could multiply identity with itself and would get all different elements of the group. But, if it was the case, could not we apply the same method to any other group (even without the assumption) and generate the whole group by the identity element? So, probably all groups would be cyclic, generated by the identity!!!