# 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.

• Sure. You end up basically re-proving Lagrange's theorem for groups of prime order, but this is certainly easier than proving Lagrange's in full generality first. Commented Feb 6, 2012 at 0:18
• A related question: math.stackexchange.com/questions/28332/… Commented Feb 6, 2012 at 4:57

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 $$G = \langle a \rangle$$.

• why was this downvoted , was it because it uses lagrange theorem or is there something wrong in the proof Commented Dec 9, 2019 at 11:47
• I know this is old. But if two groups has the same cardinality, they might have a different structure, am I wrong? Commented Aug 26, 2023 at 23:53
• @FabrizioGambelín The cyclic group generated by $a\in G$ is a subgroup of $G$. A subgroup of $G$ with the same cardinality is only $G$ itself. Commented Dec 15, 2023 at 0:38

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.

• I think this was a comment of Prof McLeman's. This is neat, by the way! Commented Feb 6, 2012 at 12:48
• @DylanMoreland: Oops, I confused the two comments! Commented Feb 6, 2012 at 13:11
• I just noticed I assumed that $x^p=1$, which is still unproven at that point. Perhaps this can be fixed, but I don't have time to figure out how right now. Commented Feb 6, 2012 at 13:22
• How you derived this $(x^a)^k \cdot (x^b)^\ell = 1^k \cdot 1^\ell$? Say, why $x^a = 1.$
– Yola
Commented Nov 14, 2016 at 5:24
• @Yola I think the hypothesis ought to be $x^a = x^b = 1$. Commented Oct 18, 2020 at 12:08

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.

• correction : ...isomorphic to $\mathbb{Z_n}$ under modular addition. Commented Nov 22, 2016 at 2:56

This is a proof of Cauchy's theorem that does not use Lagrange, it is due to James McKay. It has an uncanny similarity to the proof of Fermat's Little theorem using necklaces.

Let $G$ be a group of order $np$. Then there are $(np)^{p-1}$ solutions to $g_1g_2\dots g_p=1$ since for any values of $g_1,g_2,\dots ,g_{p-1}$ there is a unique inverse for $g_1g_2\dots g_{p-1}$. We call $S$ the set of solutions, we have asserted $|S|$ is a multiple of $p$.

Notice if $g_1,g_2\dots g_p=1$ then $g_ig_{i+1}\dots g_pg_1\dots g_{i-1}=1$ also.

Divide $S$ in rotation classes. Where $s$ is in the same class as $s'$ only if they are rotations of each other. Notice all classes have size $1$ or $p$ (this uses $p$ is prime).Therefore the number of classes of size $1$ is multiple of $p$. Since $\underbrace{1,1\dots ,1}_\text{p times}$ makes up a class of size $1$ there must be another, this provides the desired element of order $p$.

We may now use Cauchy Theorem to determine a group $G$ of order $p$ has an element $g$ of order $p$, this element generates a cyclic subgroup of order $p$. This subgroup must be $G$ so $G$ is cyclic.

• I'm struggling to see "Notice all classes have size $1$ or $p$ (this uses $p$ is prime)", could you give more detail/a hint on that part? I understand that the classes of size $1$ are the ones where the $g_i$ are all equal, and I intuit that there would be at least (and thus exactly) $p$ rotations where at least one of the $g_i$ is distinct (you can "split" the product at any $i \in \{0, \ldots, p - 1\}$ and get a different rotation), but I can't seem to prove it formally. Commented Oct 18, 2020 at 12:26
• All that we need to do is prove that if you take the element in the back and put it in the frunt (and push the other p-1 elements one position each ) then the product is still the identity, the reason for this is that $g_p$ is going to be the inverse of $g_1\dots g_{p-1}$. So this basically rotates the expression by one place, and we can do that $p$ times until we repeat the same one and they are all different rotations because $p$ is prime. I hope this helps. Commented Oct 18, 2020 at 14:28
• Thanks! Yeah I think I got so far. The only bit I'm unsure about is "they are all different rotations because $p$ is prime". How does $p$ being prime relate to the rotations being unique? Commented Oct 19, 2020 at 8:42
• suppose rotating $k$ times gives the same rotation with $k<p$ and now take an integer $a$ such that $ak$ is congruent to $1$ mod $p$. You get that the rotation is also the same after rotating $ak$ times which is the same as rotating only once. So the configuration is invariant under a single rotation which means it has only one value. If $p$ is not prime is is possible that $p$ and $k$ are not coprime so we need $p$ to be prime. Commented Oct 19, 2020 at 15:44