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If $G$ has no proper subgroup, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.

I know that since $G $is a group with no proper subgroups, $g \in G$ is not just the identity. I don't know where to go from there.

  • 1
    What's the subgroup generated by $g$? What does it mean for that subgroup to not be proper? How does its order relate to the presence of other subgroups?2011-12-15
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    This is already true (and not any harder) if we don't assume that $G$ is finite.2011-12-15
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    for $g\in G$, $\langle g\rangle$ is a cyclic subgroup. the subgroups of a cyclic subgroup of order $n$ are cyclic of order $d|n$. youre condition is then $G=\langle g\rangle$ and order of $g$ equal to $p$ for some prime $p$ (or the trivial group)2011-12-15

4 Answers 4