# Groups with $\wedge$-irreducible trivial subgroup

Suppose $G$ is a group satisfying the following condition: $$H \cap K = \{1\} \implies H = \{1\} \;\text{ or }\; K=\{1\}$$ for any two subgroups $H$, $K$, i.e. the trivial subgroup is $\wedge$-irreducible in the lattice of subgroups $L$ of $G$. I was wondering if this implies that $L$ is totally ordered. My intuition suggests me it doesn't (the condition doesn't look strong enough), but I haven't found any counterexample yet.

Consider first the case of $G$ finite. If $G$ is itself trivial, $L$ is totally ordered. Otherwise it is easy to see that it must be a $p$-group. Indeed, if $\lvert G \rvert$ has two distinct prime factors $p$ and $q$, then $G$ has two non-trivial subgroups of orders $p$ and $q$ whose intersection is trivial, which contradicts our assumption. So $\lvert G \rvert = p^n$ for some prime $p$, with $n > 0$.

Suppose $G$ is also abelian. If $G$ is cyclic, then $L$ is again totally ordered. If it is not, then $G = H \oplus K$ for some subgroups $H, K$, so we find another contradiction.

Therefore any finite counterexample must be a non-abelian $p$-group.

For the infinite case I wouldn't even know where to look for counterexamples. Does $G$ still have to be a $p$-group? It doesn't seem necessary anymore: if $G$ is not a $p$-group, then the only thing I know is that there exist two elements such that their orders are not powers of the same prime, but this doesn't imply that we can find two elements with coprime orders (from which we would find a contradiction as above). Or does it?

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Among infinite groups the infinite cyclic group $\mathbb{Z}$ is an example, and I think $(\mathbb{Q}, +)$ is another example as well. So $G$ does not have to be a $p$-group to have the property you mention. However, we can prove that if $G$ is not a $p$-group and has this property, then $G$ must be torsionfree.

Suppose that $G$ has the property that $H \cap K = 1$ implies $H = 1$ or $K = 1$ for any subgroups $H$ and $K$ of $G$.

If $G$ contains a nontrivial element of finite order, then it must contain a subgroup $H$ of prime order $p$. Then for any nontrivial subgroup $K$ we have $H \cap K \neq 1$, so in fact $H \cap K = H$ and thus $H \leq K$. Thus every element of $G$ has finite order, since the infinite cyclic group has no nontrivial finite subgroup. Hence $G$ must be a $p$-group, and $H$ is the unique subgroup of order $p$ in $G$.

Conversely, if $G$ is a $p$-group with a unique subgroup of order $p$, you can show that $G$ will have the property.

Therefore the finite groups having this property are precisely the cyclic groups of prime power order and generalized quaternion groups. For an infinite example of a $p$-group with a unique subgroup of order $p$, consider the Prüfer $p$-group $\mathbb{Z}(p^\infty)$.

Here's an infinite nonabelian example. Let $Q_{2^n}$ be the generalized quaternion group of order $2^n$, where $n \geq 3$. Now we can embed each $Q_{2^n}$ into $Q_{2^{n+1}}$, so $Q_8 \leq Q_{16} \leq Q_{32} \leq \ldots$ and we can take the union of this chain:

$$G = \bigcup_{n = 3}^\infty Q_{2^n}$$

Now $G$ is an infinite, nonabelian $2$-group with an unique subgroup of order $2$.

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This is way more than what I thought could be said about this property, thank you! By the way, $\mathbb{Z}$ is such an easy example that I wonder why I didn't think about it before. I guess in my mind every subgroup lattice had to be atomic, so I was looking for groups such that the intersection of all non-trivial subgroups is non-trivial. – Luca Bressan Sep 5 '13 at 11:21

The minimal counterexample is the quaternion group, $Q = \{ \pm 1, \pm i, \pm j, \pm k\}$. Every nontrivial subgroup contains $-1 = i^2 = j^2 = k^2$, so we have $H\cap K = \{1\} \Rightarrow H = \{1\} \lor K = \{1\}$, but $\langle i\rangle \not \subset \langle j\rangle$, and $\langle j\rangle \not\subset \langle i\rangle$.

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I see, thank you for the fast answer. Maybe this sounds stupid now, but what about the infinite case? Can an infinite counterexample be obtained from this one? I'm not familiar with this sort of things. – Luca Bressan Sep 5 '13 at 10:15
Doesn't sound stupid at all. Although I'd be very surprised if there weren't infinite counterexamples, I don't see how to obtain one from the quaternion group. Doesn't say much, because I'm not familiar with this sort of things either. I just happened to see that $Q$ is a counterexample. – Daniel Fischer Sep 5 '13 at 10:24