Show that every group of prime order is cyclic

Show that every group of prime order is cyclic.

I was given this problem for homework and I am not sure where to start. I know a solution using Lagrange's theorem, but we have not proven Lagrange's theorem yet, actually our teacher hasn't even mentioned it, so I am guessing there must be another solution. The only thing I could think of was showing that a group of prime order $p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. Would this work?

Any guidance would be appreciated.

Solution 1:

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 $x^a = x^b$, 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.

Solution 2:

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 $G = \langle a \rangle$.

Solution 3:

The answer is fairly simple once Lagrange's Theorem is quoted. We have no proper subgroups of smaller order. We only need to prove the uniqueness of the group of that size. For this note that given any element of such a group, continue to take powers of it ... This series $x^r$ has to terminate because of closure. The series also has to exhaust all the elements of the group, otherwise we will have subgroups of a smaller order.

Thus we have proven that every group of prime order is necessarily cyclic. Now every cyclic group of finite order is isomorphic to $\mathbb{Z}_n$ under modular addition, equivalently, the group of partitions of unity of order $|G|$. Thus the uniqueness is proved.