A group with 3 subgroups. prove it is cyclic

Question

Let G be a group with exactly three subgroups. Prove that G is cyclic.

So I know 2 of the subgroups: e (the identity) and G. And {e} and G are distinct. My first thought is to show that G has a generator and that's both e and G right? Not sure what/how to find the third subgroup and what to do after that.

This is what I have so far:

PROOF: Let G be a group with exactly three subgroups. Then by definition, {e} is a subgroup of G and a generator of G. Also, G is a subgroup of G and is also a generator.

Actually this also holds if $G$ has two non-trivial subgroups (so four subgroups total). See this thread. It does not hold, if $G$ has three non-trivial subgroups. The Klein four group being a counterexample. — Jyrki Lahtonen, Feb 03 '15 at 08:19
What is your definition of "generator"? It seems not to be what I would understand under this word. — Paŭlo Ebermann, Feb 03 '15 at 20:05
If any of the answers was helpful consider accepting it. Check this thread. — Aaron Maroja, Feb 10 '15 at 11:11

score 18 · Answer 1 · edited Feb 03 '15 at 07:03

18

Let $H$ be the third subgroup, so by assumption $\{e\}\ne H\ne G$. Then there exists $x\in G\setminus H$. Then $\langle x\rangle$ must be one of $\{e\}, H, G$ and it certainly is neither of the first two. (The argument can be generalized as follows: A group with a proper subgroup that contains all proper subgroups is cyclic)

edited Feb 03 '15 at 07:03

Viktor Vaughn

19,278

answered Feb 03 '15 at 06:06

Hagen von Eitzen

374,180

Wouldn't a more natural generalization of your proof be, "A group which is not covered by its proper subgroups is cyclic"? - that's even a sufficient and necessary condition for a group being cyclic. – Milo Brandt Feb 04 '15 at 04:18

score 3 · Answer 2 · answered Feb 03 '15 at 06:06

You're using the term generator a little too loosely here. The term generator means that if $g$ generates a subgroup $H$, then $H = \{g^n:n\in\Bbb Z\}$. $H$ is then denoted $\langle g\rangle$.

As for the problem at hand: suppose $g\neq e\in G$. What can you say about $\langle g\rangle$? Can there be any elements not in $\langle g\rangle$?

silvascientist · Answer 3 · 2015-02-04T04:11:21.697

Since G has only three subgroups, H has only two - $\lbrace e\rbrace$ and itself (otherwise we would have a fourth subgroup in G). So H is cyclic, i.e., $H=\langle h\rangle$ for some $h \in H$. More importantly, though, consider $g \in G$ not in H, and consider $\langle g \rangle$. Since $g \not\in H$, we have $\langle g \rangle \neq H$. What else could $\langle g \rangle$ be, then?

As an aside, to clarify what a generator is, remember that we define a generator to be an element $g$ of the group such that every element of the group is a multiple of it, i.e. $\forall x \in G, x = g^k$ for some $k \in \mathbb Z$. So $e$ cannot be a generator, since the only multiple of $e$ is $e$ itself. Similarly, $G$ cannot be a generator, since being a generator is something we define for elements, not sets.

A group with 3 subgroups. prove it is cyclic

3 Answers3