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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.

user12691
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    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
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    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

4 Answers4

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

Mikko Korhonen
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  • @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
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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. $\Box$

Arturo Magidin
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  • I don't understand why you said that g^2 ≠g and then ⟨g^2⟩=⟨g⟩ – user12691 Dec 16 '11 at 01:37
  • 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 01:45
  • @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 01:56
  • I understand $g = (g^2)^k$, but couldn't figure out how we get $g^{2k - 1} = 1$? – Ari Royce Hidayat Apr 27 '20 at 18:30
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    @AriRoyce: $(g^2)^k = g^{2k}$. Now take $g=g^{2k}$, and multiply both sides by $g^{-1}$. – Arturo Magidin Apr 27 '20 at 18:31
  • Sorry got it :D)) $g = (g^2)^k$; $1 = g^{-1} (g^2)^k$; $1 = g^{2k - 1}$. – Ari Royce Hidayat Apr 27 '20 at 18:31
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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.

Michael Hardy
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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.

Bill Dubuque
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