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I need to prove that if $G$ is an infinite group in which every non trivial proper subgroup is maximal, $G$ is simple.

I've found this: Let $G$ be a non-trivial group with no non-trivial proper subgroup. Prove that $G$ cannot be infinite group. but it says nothing about maximals. I must show that there can't be normal proper subgroups, right? Well, if every proper non trivial subgroup is maximal, it means that there can't be another proper one that contains this and it's different. In other words: $H$ is maximal if there is no other proper subgroup $K$ such that $H\subset K$ strictly. I cannot see a connection to normality of such subgroups. Which should be the way to tackle this problem?

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Use the fact that any group with no nontrivial subgroups must be finite (try to prove this for yourself if you didn't already know this):

If $G$ were not simple, it would have a nontrivial normal subgroup $N$, which must be maximal. Hence $G/N$ has no nontrivial subgroups, and is therefore finite. On the other hand, using the hypothesis, you can also show that $N$ has no nontrivial subgroups. This implies that $N$ is finite.

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Let $G$ be infinite, and suppose $G$ is not simple. Then let $N\leq G$ be a proper nontrivial normal subgroup of $G$.

If $N$ is infinite, then by the theorem you've linked to, $N$ has a nontrivial proper subgroup $H$. Then $H$ is a non-maximal nontrivial proper subgroup of $G$.

Now suppose $N$ is finite. Then $G/N$ is infinite, and thus has a nontrivial proper subgroup $H$. By the correspondence theorem, $H\cong M/N$ where $N\leq M \leq H$. Since $H$ is nontrivial and a proper subgroup of $G/N$, we have that $N\subsetneq M \subsetneq G$, and so $N$ is not a maximal subgroup of $G$.

Thus, if $G$ is infinite and non-simple, then it has a non-maximal nontrivial proper subgroup. Contrapositively, an infinite group $G$ whose proper nontrivial subgroups are maximal is simple.

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