How many elements will be there in a group so that it has $4$ subgroups of order $3$? I m thinking that a group of order $3$ will be of the form $\langle a \rangle=\{1,a,a^2=a^{-1} \}$, $ a^3=e$ therefore $a^2=a^{-1}$.
Therefore $\langle a \rangle=\{1,a,a^{-1} \}$.
Then there will be $3$ cyclic groups different then "$a$" ie "$b$", "$c$", "$d$" with no element same as any element of "$a$". So each group will have $2$ elements of order $3$. in total $4\cdot 2$ elements. Is this reasoning correct that all subgroups should be distinct?
 A: Note that $S_4$ and $A_4$ have exactly $4$ subgroups of order $3$, the Sylow $3$-subgroups. So your conclusion that the group should have $8$ elements is not valid.
A: Generalizing this a bit: from the proof (1959) of James McKay of Cauchy's Theorem (the existence of an element of prime power order $p$, if $p$ divides the order of $G$) it follows that $$\#\{g \in G: o(g)=p\} \equiv -1 \text{ mod }p$$ 
If there are $l$ subgroups of order $p$, then $l=1$ or any pair of different subgroups of order $p$ have trivial intersection (apply Lagrange's Theorem on the intersection, being a subgroup of each of the subgroups of order $p$). Hence, leaving out the identity element, there are $l(p-1)$ elements of order $p$. Combining this with the above formula yields $$l \equiv 1 \text{ mod } p$$
So in your case taking $p=3$, the minimal number of subgroups of order $3$ in a group is $4$, the next possibility would be $7$, $10$, ... etc.. As Hagen pointed out $G=C_3 \times C_3$ is the minimal example, and you can produce very large examples by looking at $G \times S$, where $S$ is a group with $3 \nmid |S|$.
