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Show that $G$ is cyclic if $G$ has at most two proper subgroup

First of all it is not a duplicate I know there is an answer here: If $G$ has only 2 proper, non-trivial subgroups then $G$ is cyclic

However, I did not understand the answer. So I will write my attempt.

My Attempt:

$\underline{\text{Case 1 : } \vert G\vert=pq, \text{where p and q are primes}}$

$\vert H\vert$ divides $pq$ and $\vert K\vert$ divides $pq$ by Lagrange Theorem. By assumption these subgroups are unique. Thus G is cyclic. Since there is at most four subgroups with orders $1, p,q,pq$.

$\underline{\text{Case 2 : } \vert G\vert=p^2, \text{where p is prime}}$

Again Lagrange implies only three subgroups of order $1,p,p^2$. A contradiction! So, in $pq$ we can assume $p \neq q$

$\underline{\text{Case 3 : } \vert G\vert=n, \text{where n is composite}}$

Since $n$ is composite we can write $n=mk$, where $1<m,k<n$. Again by Lagrange Theorem we have:

$$\vert H \vert \text{ divides } m\cdot k \text{ and } \vert K\vert \text{ divides } m\cdot k$$

These are only subgroups. Hence this implies $m$ and $k$ must be prime. Otherwise, some subgroup of order $l$, where $l$ divides $m\cdot k$. But this means we have more then $2$ subgroups. This means $n=pq$ with $p \neq q$.

Are there any mistake?

Shaun
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beingmathematician
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  • I can't follow your argument in case $1$. How do you see that $G$ is cyclic in this case? Nor can I follow your argument in Case $2$ (I didn't look at your argument for case $3$). – lulu Jul 10 '22 at 11:11
  • By Fundamental Theorem of Cyclic Groups we know there is exactly one subgroup of order d. So we know there exist subgroups of $p$ and $q$ also $1$ and $pq$ are trivial ones. So we are done – beingmathematician Jul 10 '22 at 11:12
  • If we have order $p^2$ then we have subgroups of order $1,p,p^2$ this is a contradiction since assumption says there exist at most two proper subgroup – beingmathematician Jul 10 '22 at 11:13
  • Neither of those arguments are correct. The symmetric group, $S_3$ has several subgroups of order $2$, for instance. – lulu Jul 10 '22 at 11:19
  • You seem to have forgotten that a group can contain more than one subgroup of a given order – lulu Jul 10 '22 at 11:20
  • Study the proof given in the link. It seems very clear and efficient. – lulu Jul 10 '22 at 11:21
  • But by our assumption these two subgroups are unique? – beingmathematician Jul 10 '22 at 11:23
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    Your arguments are incorrect, or at least incomplete. Study the proof in the link. – lulu Jul 10 '22 at 11:24

3 Answers3

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Here is a case-free solution, that also deals with infinite groups.

Suppose that $G$ is not cyclic, and let $X=\langle x\rangle$ be a cyclic subgroup. As $G\neq X$, there exists $y\in G\setminus X$. Now set $Y=\langle y\rangle$. If $G\neq X\cup Y$, then there is an element $z\in G\setminus(X\cup Y)$, and $\langle z\rangle$ provides us a third subgroup. Thus assume that $G=X\cup Y$.

We claim $xy$ lies in neither $X$ nor $Y$, contradicting our assumption. Notice that $y\notin X$ by construction, and $x\notin Y$ as then $X\leq Y=G$ is cyclic. So $x\in X$, $y\notin X$, so $xy\notin X$ (it lies in the coset $Xy\neq X$), and similarly $xy\notin Y$ (it lies in $xY\neq Y$). Thus $X\cup Y<G$ and we are done.

David A. Craven
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We'll use the result that a group can't be the union of two proper subgroups. (There are multiple proofs on this site. )

Take $x\not\in H_1\cup H_2$, where $H_1,H_2$ are the only two proper subgroups of $G$.

Consider $\langle x\rangle $. It's a third subgroup. Therefore $\langle x\rangle =G$.

calc ll
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  • I like that fact about unions. I had seen it before (judging by an upvote of mine on a question about it), but I forgot about it entirely. I don't know why, as it seems handy. So thank you. (+1) – Shaun Jul 10 '22 at 13:54
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The arguments are in right direction but they don't look precise. For example in case 1, what if both $|H|=|K| = p$? In case 2, I don't think the fundamental theorem even applies if there are two subgroups of order $p$. A better way will be to prove the contrapositive i.e. if the group is not cyclic then there are $3$ or more proper subgroups.

So choose $a\in G$. Consider cyclic group $\langle a\rangle$. Since $G$ is not cyclic, there is $b\in G$ such that $b\notin \langle a\rangle$. Here you get your 2nd proper subgroup $\langle b\rangle$. Now you have to produce 3rd one. If the order of $a$ (or $b$) is composite then you get the third proper subgroup as a subgroup of $\langle a\rangle$ (or $\langle b\rangle$). So assume that the order of both of them is prime. We claim that $\exists$ $c\in G$, $c \notin \langle a\rangle \cup \langle b\rangle$ which gives us our 3rd proper subgroup $\langle c\rangle$. If not then we have $\langle a\rangle \cup \langle b\rangle = G$. There are two cases.

Case (1) if the orders of both $a$ and $b$ equal to the same prime $p$ then group order is $2p - 1$ and $p$ dividing $2p - 1$ is not possible.

Case (2) if the orders are distinct primes $p \not= q$ then group order is $p + q -1$. Then $p$ divides $q -1$ and $q$ divides $p-1$ which is impossible.

Thus our claim must be true.

Note : Argument in both the cases assume that $\langle a\rangle$ and $\langle b\rangle$ share no common element other than the identity. But this is easy to see using that every non identity element in a cyclic group of prime order is a generator.

Rishi
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  • This however assumes that the group in finite. One of the answers in that other post provides a very simple argument that also covers the infinite case. – Rishi Jul 10 '22 at 13:10