I want to find all groups $G$ which have only two subgroups $G$ and $\{e\}$.
I think there can be some connection with simple groups, but it's only my intuition. I would like also to prove why these groups have only such two subgroups.
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2Here's a start: consider the generated subgroup of an element that is not the identity. That must equal to $G$. – player3236 Oct 27 '20 at 19:03
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so it is one of the possibilities? – Novice Oct 27 '20 at 19:04
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What do you mean by possibilities? – player3236 Oct 27 '20 at 19:05
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I want to find all groups $G$ s. t. they have only two subgroups: ${e}$ and $G$. The answer is generated subgroup of an element that is not the identity and nothing more? – Novice Oct 27 '20 at 19:07
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1Integers modulo a prime under addition qualify – J. W. Tanner Oct 27 '20 at 19:08
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2I mean you must consider this generated subgroup(s) for any nontrivial group $G$. Use that to reach the conclusion that $G$ must be a finite prime cyclic group. – player3236 Oct 27 '20 at 19:09
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First, notice that for $G=\lbrace e \rbrace$, then $G$ has only one subgroup, so it is not solution.
So let's suppose that $G\neq\lbrace e \rbrace$ and $G$ has only two subgroups. Then take $x \in G \setminus \lbrace e \rbrace$. The subgroup generated by $x$ must be different from $\lbrace e \rbrace$, therefore it must be equal to $G$. This means that $G$ is generated by one element : if $G$ is infinite, then $G$ is isomorphic to $\mathbb{Z}$, and $\mathbb{Z}$ has a lot of subgroups, so $G$ is not a solution. So $G$ is finite, so $G$ is cyclic. If its order is not a prime number, you can easily find several subgroups ; whereas if its order is a prime number, you cannot.
So the solutions are the cyclic groups of prime orders.
TheSilverDoe
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