Show that a group with no proper nontrivial subgroups is cyc

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Homework Statement


Show that a group with no proper nontrivial subgroups is cyclic.

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The Attempt at a Solution


If a group G has no proper nontrivial subgroups, then its only subgroups are ##\{e \}## and ##G##. Assume that G has at least two elements, and let ##a## be any element besides ##e##. Then ##a## generates a subgroup of ##G##, but ##G## has no proper nontrivial subgroups, which means that ##a## must generate ##G##, so ##G## is cyclic.

I feel that I am on the right track, but I also don't feel like I am being rigorous enough.
 
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Your proof is fine. I would just replace "##a## generates a subgroup of ##G##" by "##a## generates a non-trivial subgroup of ##G##", and observe that ##\langle a\rangle##, the subgroup generated by ##a##, cannot be proper, before stating that ##\langle a\rangle=G##.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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