Example of finite non-abelian group with nontrivial intersection of subgroups I am looking for an example of a finite non-abelian group $G$ such that the intersection of every non-trivial subgroup $H$ is another non-trivial subgroup $H_o$. Here by trivial i mean a subgroup consisting of only the identity element, $<e>$.
I have made a number of attempts at this problem and searched online, but failed to find or come up with anything helpful. Any hint, partial answer or complete solution would be very nice.
Thanks in advace.
 A: I think $Q_8$ works fine. 
Its subgroups are $$\{1\}, \{1,-1\}, \{1,-1,i,-i\},\{1,-1,j,-j\}, \{1,-1,k,-k\}, Q_8$$
According to your definition of non trivial, the non trivial subgroups are 
$$ \{1,-1\}, \{1,-1,i,-i\},\{1,-1,j,-j\}, \{1,-1,k,-k\}, Q_8$$
and their intersection is $$\{1,-1\}$$ which is a non-trivial subgroup.
A: The easiest example is the quaternion group. Since it is a group of order $8$ and has only one subgroup of order $2$, every subgroup must contain that single subgroup of order $2$.
We can show that such a group $G$ must be of prime power, and $H_0$ must be of prime order: $H_0$ must be cyclic (otherwise we take one nontrivial element and use its cyclic subgroup) and thus of prime order. If the order of $G$ is $p^nq$ where $p \nmid q$ and $q \neq 1$, then we can find a subgroup of order dividing $q$ which must have trivial intersection with $H_0$, contradicting the definition of $H_0$. Thus the order of $G$ must be $p^m$. Since every group of order $p$ or $p^2$ is abelian, we need $m \geq 3$. So the first place to look is at groups of order $2^3=8$ ($p=2$,$m=3$), and indeed the quaternion group works.
A: I think you are looking for quaternion group of order 8 ($Q_8$). Its non-trivial subgroups are:
$
H_1=\{ 1, -1 \}\\H_2=\{1, -1, i, -i\}\\H_3=\{1, -1, j, -j\}\\H_4=\{1,-1, k, -k\}
$
Observe that any two non-trivial subsgroups intersect non-trivially, (the intersection is $\{1, -1\}$).
A: A finite group $G$ with the property that the intersection of any two nontrivial subgroups is nontrivial is a $p$-group for some prime order, and either cyclic, or $p=2$ and it's generalized quaternion.
Indeed, that $G$ is a $p$-group is clear, and the property of having a single element forces the given structure, by an old result of Burnside.
For $G$ infinite, we still can argue that $G$ is torsion-free, or is a $p$-group for some $p$, but there might be exotic (= not locally finite) examples (I don't know if there's a torsion-free one). 
