I want someone to check my proof. Suppose every proper subgroup is cyclic,but G is not cyclic. Since subgroup is cyclic,It generator must exist in subgroup. Then all element in subgroup exist in G;it included the generator of subgroup. then G has generator and cyclic contradict my assumetion therefore if every proper subgroup is cyclic G is cyclic
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Try $S_3$. Only proper subgroups have order $1,2$ or $3$. – lulu Nov 01 '16 at 19:59
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You cannot deduce "then G has generator...", because you only know that a generator of a subgroup exists in $G$. Did you try looking for some examples? – Wojowu Nov 01 '16 at 20:00