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Is the following true?

If $G$ has two proper, non-trivial subgroups then $G$ is cyclic.

ROBINSON
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  • Did you mean two proper non-trivial subgroups? – Tobias Kildetoft May 01 '13 at 16:31
  • yeah, I mean two proper non-trivial subgroup here. – ROBINSON May 01 '13 at 16:36
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    @AKASNIL: In that case you should go back and edit the question right away, since that's not what you asked and lots of people are answering the question you actually asked. – Pete L. Clark May 01 '13 at 16:46
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    This is more of a general comment on all the responses thus far. In the initial set of comments on the question, the OP stated that he was referring to a group with exactly two proper, non-trivial subgroups. I assumed he meant exactly two subgroups other than ${1}$ and $G$. Am I missing something here? This possibility is all the more likely now that I have recently celebrated yet another birthday. – Chris Leary May 01 '13 at 16:53
  • Please edit the title of your question too if this is possible. – Stefan Smith May 01 '13 at 17:29
  • @rschwieb: This is in reply to a comment on your deleted answer. I never said that anyone had "an inability to reason correctly"; I said that two people had "made a basic logical error". There is no blanket statement about people's cognitive or logical abilities there. So far as I know, all human beings make basic logical errors from time to time. I am quite sure that I do... – Pete L. Clark May 01 '13 at 17:37
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    @Pete - I've always believed that if you try to reason you will make logical errors from time to time. Some of the ones I have made have astounded me once I realized them. That's pretty much the price we pay for being human. – Chris Leary May 03 '13 at 15:45

4 Answers4

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Let $H_1$ and $H_2$ be the two non-trivial proper subgroups of the given group $G$. I claim that $G$ is not the union $H_1\cup H_2$. If one of the subgroups is contained in the other, then this is trivially true. Otherwise there exist elements belonging to one subgroup but not the other. Let $h_1\in H_1\setminus H_2$ and $h_2\in H_2\setminus H_1$. What about $g=h_1h_2$? If it belongs to $H_1$, then so does $h_2$. If it belongs to $H_2$, then so does $h_1$. In either case we contradict our assumptions, so we have to conclude that $g\notin H_1\cup H_2$.

So we know that there exists an element $g\in G$, $g\notin H_1\cup H_2$. What is the subgroup generated by $g$? Can't be either $H_1$ or $H_2$, so it has to be all of $G$. Ergo, $G$ is cyclic.

Jyrki Lahtonen
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    Very nice and elementary answer. – Tobias Kildetoft May 01 '13 at 17:00
  • Complete, yes. Not so sure about clean. If $G$ were finite, it would be easier to prove that $G\neq H_1\cup H_2$ (Lagrange is all you need). I didn't see a nice way of covering the infinite groups in the same argument, so I resorted to the uglier way of picking those $h_1,h_2$ :-( – Jyrki Lahtonen May 01 '13 at 19:50
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    But the fact that the union of two subgroups is not a subgroup unless one is contained in the other is standard knowledge (or should be). – Tobias Kildetoft May 02 '13 at 00:45
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    Unlike Jyrki, I think his argument using $h_1$ and $h_2$ is prettier than the one using Lagrange's theorem. Not only does it work for infinite groups, but it uses only information that is even more basic than Lagrange's theorem. – Andreas Blass May 02 '13 at 00:50
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    +1: With respect to some esthetic that I would have trouble enunciating, this is clearly the best possible answer. – Pete L. Clark May 02 '13 at 01:47
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First note that if $G$ does not have finite order, then it does not have a finite number of subgroups, so we can assume that $G$ is finite (see the comment by Pete Clark).

Note that if $3$ distinct primes divides the order of the group, then the group has at least $3$ proper non-trivial subgroups.

So $|G| = p^nq^m$ with $p$ and $q$ primes. Now, if either $n$ or $m$ is greater than or equal to $4$, then the corresponding Sylow subgroup has too many subgroups. Also, if either if at least $2$ and the other is not $0$, we again get too many subgroups.

We are left with either $|G| = p^3$ or $|G| = pq$. In both cases the cyclic group of that order will satisfy the conditions, and we wish to show that these are the only ones (since the cyclic group of order $p^2$ has too few subgroups, and the non-cyclic one has too many).

If $|G| = pq$ and $G$ is not cyclic, then $G$ is not abelian, and thus has more than one Sylow subgroup for either $p$ or $q$, giving us too many subgroups.

If $|G| = p^3$ then $G$ has at least one subgroup of order $p$ and one of order $p^2$. But if $G$ is not cyclic, it has more than one maximal subgroup, which gives us at least two of order $p^2$, again resulting in too many subgroups.

Tobias Kildetoft
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  • Assuming $G$ is finite? – Metin Y. May 01 '13 at 16:45
  • @MetinY. Right, no infinite group can have this property. – Tobias Kildetoft May 01 '13 at 16:47
  • This is a nice answer. One minor point: you seem to be assuming that the group is finite, which the OP did not say (although s/he didn't say other things as well...) and in any case is not necessary. But an infinite group either contains an element of infinite order or arbitrarily large finite subgroups, so this is no problem. I do think you should put this in the answer, though. – Pete L. Clark May 01 '13 at 16:53
  • @PeteL.Clark Thanks, edited to clarify. – Tobias Kildetoft May 01 '13 at 16:55
  • @Pete: Dear Pete, As an aside: do you know a slick proof that an infinite gp. of bounded exponent has arbitrarily large finite subgroups? (When I think about this in general I get a bit confused by Burnside problem issues, although it's very easy to deal with the particular case of having only two proper subgroups!) ... [coming back a few minutes later] actually this is not true, the Tarksi monster gives a counterexample. Regards, – Matt E May 01 '13 at 18:02
  • Ahh, that does make the original statement incorrect (though any infinite group will of course still have an infinite number of different subgroups). – Tobias Kildetoft May 01 '13 at 18:04
  • @Matt: Whoops, I must have made a mistake (perhaps a "basic logical error", although I suspect it was worse than that!). The Tarski monster groups are so strange that I haven't yet hammered them into my intuition. – Pete L. Clark May 01 '13 at 18:26
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    Let me try again: I claim that every infinite group $G$ contains infinitely many subgroups. Indeed, this is clear if it contains an element of infinite order. If not, it contains infinitely many elements of finite order, hence infinitely many finite cyclic subgroups (but they may well all have the same order, despite the fact that my intuition still suggests that this is impossible). – Pete L. Clark May 01 '13 at 18:27
  • @PeteL.Clark What about the group $(\mathbb{Z}^{\geq0}, \oplus)$ where $\oplus$ is the usual XOR/nim-addition operator? Every element of this group has order 2, so there are infinitely many distinct copies of $\mathbb{Z}/2\mathbb{Z}$ in it. (This in response to the parenthetical, of course, not the claim itself which looks fine) – Steven Stadnicki May 02 '13 at 01:13
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    @StevenStadnicki That group does have arbitrarily large finite subgroups. As mentioned previously, an example where all the proper non-trivial subgroups have the same (finite) order is given by the Tarski monster. – Tobias Kildetoft May 02 '13 at 01:21
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Since $G$ has proper non-trivial subgroups $\exists~a~(\neq e)\in G.$

  • Case $1$: $G=(a):$ Nothing left to prove.

  • Case $2$: $(a)$ is a non-trivial proper subgroup of $G:$ Choose $b\in G-(a).$

    • Case $2.1:$ $G=(b):$ Nothing left to prove.

    • Case $2.2:$ $(b)$ is also a non-trivial proper subgroup of $G:$

      • Case $2.2.1:$ $(a)\cup(b)=G,$ a subgroup of $G.$

        Consequently either $(a)\subset(b)$ or $(b)\subset(a).$

      • Case $2.2.2:$ $\exists~c\in G-(a)\cup(b).$

        Since $G$ has only two proper subgroups $G=(c).$

Sugata Adhya
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  • @Adhya : I like your answer, but you lost me in the middle of Case 2. Do you mean "Since $G$ has only two proper subgroups..."? And why do you state that $G$ is a subgroup of $G$? Is this a typo? – Stefan Smith May 01 '13 at 21:42
  • @Adhya : If $(a) \cup (b)$ is a subgroup of $G$, does it follow that $(a) \subset (b)$ or $(b) \subset (a)$? I don't know much group theory. I upvoted your answer because it is so elegant. – Stefan Smith May 01 '13 at 21:52
  • @StefanSmith: See http://groupprops.subwiki.org/wiki/Union_of_two_subgroups_is_not_a_subgroup_unless_they_are_comparable. – Sugata Adhya May 02 '13 at 00:25
  • @Adhya : Thanks for the link. I read the proof, which should be quite simple, and I took a survey posted there and complained about the "tabular method" of proof they used there, which I didn't care for. (I apologize if you happen to be the creator of that page/proof, and I respect the effort you put into it) – Stefan Smith May 02 '13 at 00:50
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Llet $|G| = n$. Suppose $a,b$ be two nonidentity elements of $G$. Now consider $\langle a \rangle$ and $\langle b \rangle$. If $G$ is commutative then it is easy to see that $\langle ab \rangle$ is a cyclic group other than $\langle a \rangle$ and $\langle b \rangle$, which leads to a contradiction, so one of $a$ and $b$ must be of order $n$, so $G$ is cyclic. If $G$ is non-commutative then it can't be cyclic, so we are done.

Alex M.
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Anjan Samanta
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  • thank u Mr. Alex. I don't know how to use mathjax. – Anjan Samanta Oct 16 '17 at 08:35
  • I fail to follow your logic. If $G$ is commutative it can easily happen that $\langle a\rangle$, $\langle b\rangle$ and $\langle ab\rangle$ are all the same subgroup. And in the non-commutative case the contrapositive claim would be to prove that the group has at least three proper subgroup, so you are not done in that case either. – Jyrki Lahtonen Oct 16 '17 at 14:35
  • sorry my bad.if b is the inverse of 'a' then they can't be distinct.And the second case isn't obvious.I have done several mistakes. – Anjan Samanta Oct 16 '17 at 18:25