All finite groups $G$ which have $|G|-1$ cyclic subgroups and $|G$| is not divisible by $4$. 
Find all finite groups $G$ which have the properties:
  (i) $|G|$ is not divisible by $4$;
  (ii) $G$ has exactly $|G|-1$ cyclic subgroups.   

I observed that $\mathbb{Z_3}$ works and I also figured out that if $|G|>3$, then $G$ cannot be cyclic because it wouldn't satisfy (ii).
I also noticed tried considering the cyclic subgroups generated by each element, hoping to obtain something related to (ii), but I wasn't successful.
 A: Tarnauceanu proved in $2015$ that a
finite group $G$ has $|G| − 1$ cyclic subgroups if and only if $G$ is either $C_3,C_4, S_3$ or $D_8$. Here is the proof, which is half of a page.
Perhaps it is interesting to note the following open problem. Let $c(G)$ denote the set of all cyclic subgroups of $G$.
Open Problem: Describe the finite groups $G$ satisfying $|c(G)| = |G| − r$ , where $2 ≤ r ≤ |G| − 1$.
A: To have $|G|-1$ cyclic subgroups, every element of $G$ except one of them must generate a unique subgroup. There will be two elements $a\neq b$ that generate the same subgroup.
So except for $a$ and $b$, every element is its own inverse. And $b$ must equal $a^{-1}$.
The order of $a$ must be $3$, $4$, or $6$, or else $\langle a\rangle$ has additional generators besides $a$ and $a^{-1}$. But the order of $a$ can't be $4$ given the condition about $4$. And it can't be $6$ or else there are too many elements that are not their own inverses. So the order of $a$ is $3$.
So we have a group with exactly one order-3 cyclic subgroup $\{e,a,a^{-1}\}$, and all other elements have order $2$. Therefore the order of $G$ is $2^n3$. But $4$ doesn't divide $|G|$, so either $|G|=3$ or $|G|=6$. The only possibilities are $C_3$, $C_6$, and $S_3$. But again, $C_6$ has too many elements that are not their own inverses. The other two options check out as meeting the conditions.
