Concerning Groups having the property that intersection of any two non-trivial subgroups is non-trivial The group of rational numbers  $(\mathbb Q,+)$ has an interesting property , that the intersection of any two non-trivial subgroups of this group is non-trivial . Let us call this property the " non-trvial intersection property " or NIP in short . Now it is easy to see that this NIP property is invariant under group isomorphism , so if $G$ is a group having NIP , then $G$ cannot be isomorphic with $H \times K$ (because if $|H|,|K|>1$ , then $\{e_H\} \times K , H\times \{e_K\}$ are non-trivial subgroups of $H \times K$ with trivial intersection ) for any groups $H$ and $K$ . 
I am looking for more examples of groups having NIP , does there exist infinitely many non-isomorphic such groups ? Also ,  have this kind of groups been studied ? Any reference or link will also be very helpful . Thanks in advance 
NOTE : All groups considered are to be meant  with more than one element 
 A: You are asking for groups in which every non-trivial subgroup is essential.
We can classify all abelian groups with this property, as follows.
Embed $G$ in its injective envelope $D$, which is a divisible group that is an essential extension of $G$.  If $H,K$ are non-trivial subgroups of $D$, then $H\cap G$ and $K\cap G$ are non-trivial subgroups of $G$ (here we use that it is an essential extension).  But then their intersection $(H\cap G)\cap (K\cap G)$ is nontrivial, so $H\cap K$ is non-trivial.
So $D$ is also an abelian group with NIP.  Since $D$ is divisible, by the structure theorem it is a direct product of copies of $\mathbb{Q}$ and Prüfer groups .  Since a group with NIP cannot be a direct product, $D$ is either equal to $\mathbb{Q}$, or a Prüfer group.
So if $G$ is torsion-free, it is a subgroup of $\mathbb{Q}$.  If it has torsion, then it is a subgroup of a Prüfer group, which means that it is either cyclic of prime power order, or a Prüfer group.
The non-abelian case appears to be substantially more difficult.  As Jyrki Lahtonen points out, the quaternion group $Q_8$ is a finite example; it would be nice to know if there are infinite examples.
One observation: a group $G$ with NIP that has a torsion element must have a unique minimal subgroup of order $p$.  I believe that it is known that such a $G$ must be abelian if $p>2$, but I don't know if there is a simple proof of this, or a characterization of the non-abelian $2$-groups with unique minimal subgroup. (studiosus and Jyrki mention some infinite classes of examples)
A: The groups ${\mathbb Z}_{p^{\infty}}$  with $p$ prime have this property, and there infinitely many of these - one for each prime. In fact they have a unique minimal nontrivial subgroup, which is cyclic of order $p$. Although the groups are infinite, all of their proper subgroups are finite cyclic $p$-groups.
You could define ${\mathbb Z}_{p^{\infty}}$ to be the multiplicative group of all complex $p^k$-th roots of unity for some $k$, where $p$ is a fixed prime.
A: There are also (infinitely many) finitely generated nonabelian groups satisfying NIP. Namely, A.Olshansky constructed (for every sufficiently large prime number $p$) a finitely generated group, which I will denote $G_p$, satisfying the following properties:


*

*$G_p$ is torsion-free. 

*$G_p$ contains a central infinite cyclic subgroup $Z< G_p$ such that the quotient group $G_p/Z$ is infinite and each nontrivial element of $G_p/Z$ has order $p$.  
In particular, any two nontrivial subgroups $A, B < G_p$ have nontrivial intersection with $Z$ and, hence, $A\cap B$ is nontrivial. 
A. Olshansky, "Geometry of defining relations in groups", Springer Verlag, 1991. 
A: As examples of finite groups with this property let me add the (generalized) quaternion groups. The quaternion group $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ occurs often enough in introductory courses. It has the given property because $-1$ is the only element of order two, and thus is contained in every non-trivial subgroup.
The generalied version $Q_{2^n}, n>3,$ is defined by 
$$
Q_{2^n}=\langle a,b\mid a^{2^{n-2}}=b^2, b^4=1, bab^{-1}=a^{2^{n-2}+1}\rangle.
$$
It is easier for me to think of this as the group generated by the complex matrices
$$
A=\left(\begin{array}{cc}\zeta&0\\0&\zeta^{-1}\end{array}\right),
\qquad
B=\left(\begin{array}{cc}0&1\\-1&0\end{array}\right),
$$
where $\zeta$ is a primitive root of unity of order $2^{n-1}$.
Again, the element $-I_2=B^2=A^{2^{n-2}}$ is the only element of order two, and is thus contained in all the non-trivial subgroups.
