Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $G$ has no proper subgroup, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.

I know that since $G $is a group with no proper subgroups, $g \in G$ is not just the identity. I don't know where to go from there.

share|cite|improve this question
What's the subgroup generated by $g$? What does it mean for that subgroup to not be proper? How does its order relate to the presence of other subgroups? – Jyrki Lahtonen Dec 15 '11 at 22:05
This is already true (and not any harder) if we don't assume that $G$ is finite. – Justin Campbell Dec 15 '11 at 22:08
for $g\in G$, $\langle g\rangle$ is a cyclic subgroup. the subgroups of a cyclic subgroup of order $n$ are cyclic of order $d|n$. youre condition is then $G=\langle g\rangle$ and order of $g$ equal to $p$ for some prime $p$ (or the trivial group) – yoyo Dec 15 '11 at 22:40

Examine the cyclic subgroup generated by some $g \in G$, where $g$ is not the identity.

share|cite|improve this answer
@user12691: $\langle g \rangle$ is the group generated by $g$, so it must contain the identity element $1$ to be a group. It contains all the powers of $g$. So $\langle g \rangle = \{ 1, g, g^2, \cdots, g^{n-1} \}$, because we're talking about finite groups here (and so the order of $g$ is finite, in this case $n$). – Mikko Korhonen Dec 15 '11 at 22:37

You don't need to assume $G$ is finite.

Proposition. If $G$ is a group which has no nontrivial proper subgroups, then either $G$ is the trivial group or $G$ is cyclic of prime order.

Proof. If $G$ is the trivial group, we are done. If $G$ is not the trivial group, let $g\in G$ be any element other than the identity. Then $\langle g\rangle$ is a nontrivial subgroup of $G$, and therefore must equal all of $G$ by hypothesis. Thus, $G$ is cyclic.

If $g^2=1$, then $\langle g\rangle =\{1,g\} = G$, so $G$ is cyclic of order $2$ and we are done. If $g^2\neq 1$, then $\langle g^2\rangle$ is a nontrivial subgroup of $G$, so $G=\langle g^2\rangle = \langle g\rangle$, hence there exists $k$ such that $g = (g^2)^k$. Thus, $g^{2k-1}=1$, which proves that $g$ is of finite order. Thus, $G$ is finite cyclic.

Let $n$ be the order of $g$. If $a|n$, $0\lt a\lt n$, then $\langle g^a\rangle = G$ (since it is a nontrivial subgroup). Therefore, $g\in \langle g^a\rangle$, so there exists b such that $g = (g^{a})^b = g^{ab}$. Therefore, $g^{ab-1} = 1$, so $n|ab-1$. Since $a|n$, then $a|ab-1$, hence $a|-1$, so $a=\pm 1$.

That is, the only divisors of $n$ are $\pm 1$ and $\pm n$, so $n$ is prime. $\Box$

share|cite|improve this answer
I don't understand why you said that g^2 ≠g and then ⟨g^2⟩=⟨g⟩ – user12691 Dec 16 '11 at 1:37
That should be $g^2\neq 1$; if $g^2\neq 1$, then $\langle g^2\rangle$ is a subgroup of $G$ that is not trivial, so it must equal $G$; but $G$ already equals $\langle g\rangle$, so $\langle g^2\rangle = \langle g \rangle$. – Arturo Magidin Dec 16 '11 at 1:45
@user12691 Note that I am not asserting that $g^2\neq 1$ necessarily is true. Rather, either $g^2=1$ or $g^2\neq 1$ (by the law of the excluded middle). If $g^2=1$, then $G$ is cyclic of order $2$ and we are done, and if $g^2\neq 1$, then we continue the argument. – Arturo Magidin Dec 16 '11 at 1:56

In view of the original poster's request for further explanation after an answer that got five up-votes, here are further comments.

Let $g$ differ from the identity $e$. Look at $g, g^2, g^3, \ldots$. Since the group is finite, eventually you reach $g^m=\text{some earlier term in this sequence}= g^\ell$. So $\ell<m$. Since $g^m=g^\ell$, you get $g^{m-1}=g^{\ell-1}$ unless $\ell=0$, and that means the $m$th term wasn't the first term equal to some earlier term; the $(m-1)$th term is an earlier one. So if it's the first one, then $\ell=0$. So the sequence is $g, g^2, g^3, \ldots,g^{m-2},g^{m-1},g^m$, and the last term is $e$. This is then a subgroup. But there are no proper subgroups besides the trivial one, so you've got the whole group, and $m=n$.

That gets you a cyclic group; now you need to prove that $n$ is prime. Suppose it's not, so that $n=jk$ and $j, k$ are smaller numbers than $n$. Then $g^k, g^{2k}, g^{3k},\ldots,g^{jk}=e$ is a subgroup. But there are no proper subegroups, so the assumption that $n$ is not prime is refuted.

share|cite|improve this answer
The 5 up-voted response wasn't actually an answer. Thank you for your help. – user12691 Dec 16 '11 at 2:26
@user12691: It was a pointed hint at how you could answer your own question. – Arturo Magidin Dec 16 '11 at 4:26

HINT $\ $ Any noncyclic group has a proper subgroup generated by any non-identity element. Any infinite cyclic group $\rm\:\left<g\right>\:$ has the proper subgroup $\rm\:\left<g^2 \right>\:.\:$ A finite cyclic group $\rm\:\left<g\right>\:$ of composite order $\rm\:nk\:,\ n,k > 1\:,\:$ has proper subgroup $\rm\:\left<g^n\right>\:.\:$ What remains are cyclic groups of prime order.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.