# Inclusion of subgroups implies the group is cyclic

Let $G$ be a finite group such that for any two subgroups $H_{1}$ and $H_{2}$ of $G$ we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. Why this implies $G$ is a cyclic group?

Ah. I think this works: Suppose $g \in G$ then $G = \langle g \rangle$. Suppose otherwise, then we can find $z$ such that $z \not \in \langle g \rangle$. Now let $H_{1}=\langle z \rangle$ and $H_{2}=\langle g \rangle$. By assumption we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. In both cases we get the contradiction $z \in \langle g \rangle$.

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@Dan Petersen: I think you mean't we don't get this if $\langle g \rangle \subseteq \langle z \rangle$ right? how do we fix this? –  user8079 Mar 11 '11 at 8:51
Hint: Under the given conditions, the group must have exactly one maximal proper subgroup. Now consider an element not contained in this subgroup. –  Tobias Kildetoft Mar 11 '11 at 9:35

It might be worth noting that you do need $G$ to be finite; otherwise, the claim is not true. For example, the Prüfer $p$-group has the property that given any two elements $a$ and $b$, either $\langle a\rangle \subseteq \langle b\rangle$ or $\langle b\rangle \subseteq \langle a\rangle$, but the group itself is not cyclic.

So, we want to prove:

Suppose $G$ is a finite group such that for all $a,b\in G$, either $\langle a\rangle\subseteq \langle b\rangle$ or else $\langle b\rangle \subseteq \langle a\rangle$. Then $G$ is cyclic.

Your "I think this works" is very badly phrased. You cannot begin by assuming that $G=\langle g\rangle$, because this is tantamount to assuming that $G$ is cyclic; but this is what you want to prove! You don't get to assume it to begin with. At best, you should say: "If $G=\langle g\rangle$, then we're done; otherwise, let..."

The argument works reasonably well: suppose $G=\{g_1,\ldots,g_n\}$ (remember, $G$ has to be finite). Let $a_1=g_1$. Now, consider $g_2$; either $g_2\in\langle a\rangle$ or $\langle a\rangle\subseteq \langle g_2\rangle$. In the former case, let $a_2=g_1$; in the latter, let $a_2=g_2$.

Assume you have already defined $a_1,\ldots,a_{\ell}$, with $1\leq \ell\lt n$. Consider $g_{\ell+1}$. If $g_{\ell+1}\in\langle a_{\ell}\rangle$, let $a_{\ell+1}=a_{\ell}$. Otherwise, we have $\langle a_{\ell}\rangle \subseteq \langle g_{\ell+1}\rangle$; then set $a_{\ell+1}=g_{\ell+1}$.

Now, I claim that $G=\langle a_{n}\rangle$. Indeed, note that by construction, $g_k\in\langle a_{k}\rangle\subseteq \langle a_{k+1}\rangle$. Therefore, $g_1,\ldots,g_k\in\langle a_{k}\rangle$. In particular, $G=\{g_1,\ldots,g_n\}\subseteq \langle a_n\rangle \subseteq G$, giving equality. Thus, $G$ is cyclic.

(There are of course easier ways of doing this; but this follows the intuitive idea of your argument, only written up more formally).

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from the first draft, the OP's question has included the hypothesis of finiteness of $G$. –  Pete L. Clark Mar 11 '11 at 19:29
@Pete: Hmmm... Quite so. I wonder what momentary blindness struck me. –  Arturo Magidin Mar 11 '11 at 19:33
Well, it's nothing that I haven't been afflicted with myself...Anyway, it's good that you point out that the result does not extend to infinite groups. –  Pete L. Clark Mar 11 '11 at 21:48
Another (marginally different) proof: Consider a maximal cyclic subgroup $H$ of $G$ (one exists since $G$ is finite). If $g \in G$ then the assumption implies either that $\langle g \rangle \subset H$ or else that $H \subset \langle g \rangle$. In the latter case, we have equality, by maximality of $H$, and so in all cases $g \in H$. Thus $H = G$, and so $G$ is cyclic.
Let $a$, $b$ be two random elements. Consider the subgroup $\langle a \rangle$ and $\langle b \rangle$, I think you know how to do the rest.
the symbol for $\not\in$ is \not\in. The rest should be ok. –  Kerry Mar 11 '11 at 8:15