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


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)

  • $\begingroup$ Ah I see ... yes , I noticed that if $G$ has NIP then any nontrivial subgroup of $G$ also has NIP . That Prufer group thing you mentioned , as I don't know yet , I will have to study taking time . That question you raise , whether all torsion-free abelian NIP groups arise as subgroups of $\mathbb Q$ is very interesting ! And I am trying to crack the last divisible group comment of yours . It is a resourceful answer . Please do feel free to add anything more ... Thanks $\endgroup$ – user228168 Jul 3 '16 at 10:08
  • $\begingroup$ Is any divisible group a vector space over $\mathbb Q$ ? I don't quite see it ; I know that rational powers can be defined but what about uniqely defining it ? I know that torsion-free divisible groups can be made into such vector space . Is $D$ torsion-free ? $\endgroup$ – user228168 Jul 3 '16 at 10:14
  • $\begingroup$ @SaunDev I forgot to mention that $D$ is also torsion-free. Any torsion-free, divisible group is a vector space over $\mathbb{Q}$. Actually, there is a nice fact that any divisible group is a direct sum of copies of $\mathbb{Q}$ and the Prüfer groups. $\endgroup$ – Slade Jul 3 '16 at 10:18
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    $\begingroup$ Are you saying that a finite group with this property is cyclic of prime power order? How does $Q_8$ fit into this picture? +1 of course, for the nice stuff. $\endgroup$ – Jyrki Lahtonen Jul 6 '16 at 5:43
  • $\begingroup$ @JyrkiLahtonen You're right, I mis-applied a result that was meant for abelian groups only. It seems that there may be other examples that are $2$-groups. Anyway, I've cleaned up the argument. $\endgroup$ – Slade Jul 7 '16 at 21:55

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.

  • $\begingroup$ I see , thank you . I didn't know about these groups . Does this have a name ? $\endgroup$ – user228168 Jul 3 '16 at 9:52

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:

  1. $G_p$ is torsion-free.

  2. $G_p$ 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.


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.


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