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Characterize all the groups $G$with the following property:

There is a proper subgroup $H$ of $G$ such that $\forall S$ proper subgroup of $G$, $S \subset H$.

I am pretty lost with this exercise. If $G$ is a group such that every element $g \neq e$ generates $G$, then $G$ has the property: since $G$ has no proper subgroups, then $G$ trivially verifies the property.

If not, then it is clear that $H$ is a maximal subgroup of $G$. I don't know what to do next, any suggestions would be appreciated.

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    $\begingroup$ I would not say that the case of a group generated by any one nontrivial element satisfies due to "no proper subgroups" but rather due to the trivial subgroup being maximal. $\endgroup$
    – hardmath
    Sep 21, 2014 at 18:29

2 Answers 2

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Hint: Assume you have such a group and consider an element $g \in G\setminus H$. What is $\langle g \rangle$?

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  • $\begingroup$ Well, then $<g>$ has to be $G$, because if not, then $<g>$ would be a proper subgroup of $G$, so $g \in <g> \subset H$. So $H \cup \{g\}=G$ for any $g \in G \setminus H$, but I don't see how this condition characterizes all the groups which satisfy the property. $\endgroup$
    – user100106
    Sep 21, 2014 at 21:09
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    $\begingroup$ Well, you have infered that the group is cyclic (i.e. generated by one element). And there are only countably many types of cyclic groups: namely, up to iso, $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$, where $n \in \mathbb{N}$. It is not difficult to see that for $\mathbb{Z}$ the condition does not hold (which are the maximal subgroups?). Try to determine for which $n$ the condition holds in $\mathbb{Z}/n\mathbb{Z}$. Hint: It depends on the prime factorization of $n$. $\endgroup$
    – Pavel Čoupek
    Sep 21, 2014 at 21:15
  • $\begingroup$ @PavelC I was trying to solve this exercise and for the case $G=\mathbb Z_n$ I am having some problems trying to determine for which $n$ the property holds. If $n$ is prime then the property trivially holds, but I couldn't figure out what happens if $n$ is a composite number, could you help me with that? $\endgroup$
    – user16924
    Dec 1, 2014 at 4:10
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    $\begingroup$ @user16924: Consider two separated cases: Suppose there exist two different primes $p, q$ dividing $n$. Then one can show that there are at least two maximal subgroups of $\mathbb{Z}_n$, so the property does not hold for $\mathbb{Z}_n$. As for the other case, we have that $n=p^k$ for some prime $p$ and some $k>0$ (let us not consider the trivial group). Then it could be verified that the property holds. $\endgroup$
    – Pavel Čoupek
    Dec 1, 2014 at 16:29
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Let $g$ be an element of $G$ which is not in $H$, since $\langle g\rangle$ not contained in $H$, we must have $\langle g\rangle=G$.

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