4
$\begingroup$

Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.

I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.

$\endgroup$
  • $\begingroup$ Do you know the Sylow theorems? $\endgroup$ – abcdef Apr 4 '15 at 16:36
  • $\begingroup$ but how i apply sylow theorems $\endgroup$ – Yogi Apr 4 '15 at 16:47
  • 1
    $\begingroup$ Sylow theory is preety over-powered for this problem $\endgroup$ – Arpit Kansal Apr 4 '15 at 16:50
  • 1
    $\begingroup$ Note that this is exercise 34 in supplementary exercises to chapters 1-4 in Gallian's Contemporary Abstract Algebra. $\endgroup$ – a student Dec 6 '15 at 6:51
  • $\begingroup$ That's what brought me to this post, @astudent :) $\endgroup$ – Shaun Jan 27 at 13:28
5
$\begingroup$

Claim:$G$ is cyclic.

Proof: If not, then there must be two distinct elements $a$ and $b$ that generate two distinct cyclic subgroups. Thus, $G$ must be cyclic.
From here, we know all cyclic groups are isomorphic to $\mathbb{Z}_n$, where $n = |G|$. At this point, we are done: For all cyclic groups $H$, there will exist exactly $1$ cyclic subgroup for each divisor of $|H|$. Thus, $|G| = p^2$

$\endgroup$
3
$\begingroup$

The idea is to first show that $G$ is cyclic and then use this fact to deduce that $|G|=p^2$ where $p$ is prime.

To show that $G$ is cyclic let $G$ be a group with exactly one non trivial subgroup. Then $G$ contains at least two (different) non trivial elements $a,b \in G $. Certainly, $\langle a \rangle $ is a non trivial subgroup of $G$. If $\langle a \rangle = G$ we're done since $G$ is cyclic.

Otherwise note that we must have $\langle a \rangle = \langle b \rangle$ since there is only one non trivial subgroup. Let $c$ be any element in $G \setminus \langle a \rangle $. Then $\langle c \rangle = G$ since we cannot have $\langle c \rangle = \langle a \rangle$. Hence $G$ is cyclic.

A cyclic group has a subgroup for every divisor of the order of the group. Hence we can only have one divisor of $|G|$, that is, $|G| = p^2$ or $|G| = p$ but the latter is not possible since there is one proper subgroup.

$\endgroup$
2
$\begingroup$

You're on the right track - your idea of considering orders of members of the group is good. A slightly more general idea that resolves this is to, rather than the orders, consider, for any element $x\in G$, the subgroup generated by $x$, namely $\{1,x,x^2,\ldots\}$.

For any non-identity element, this subgroup is non-trivial. There are two non-trivial subgroups of $G$ - being $G$ itself and a proper subgroup which we'll call $G'$. Now, if we choose some element $x$ of $G$ which is not in $G'$, then it follows that the subgroup generated by $x$ must not be a subset of $G'$ - and the only subgroup which isn't is $G$ itself. This establishes that such an $x$ generates the whole group - meaning the group is cyclic.

Given that you've already proven that $x$ would have order $p^n$ for some $n$, you just need to prove that if it has order $p$, then it has no non-trivial proper subgroups and if it has order $p^n$ for $n>2$, then the subgroups generated by $x^p$ and $x^{p^2}$ are distinct non-trivial proper subgroups. This leaves that $x$ has order $p^2$ and generates the group has order $p^2$ and is cyclic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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