Existence of a non-trivial subgroup related by containment with the rest of subgroups Let $G$ be a finite group. I am interested in finite groups having a non-trivial subgroup $H$ $(H\neq \{e\}, G$) such that $H' \subseteq H$ or $H \subseteq H'$ for every subgroup $H'\neq H.$ That is, there is some non-trivial subgroup $H$ which contains or it is contained in the rest of subgroups.
For instance, if we assume that $G$ is finite and cyclic, this condition is satisfied only if $G \cong \mathbb{Z}_{p^k}$ for $p$ prime. If $G \cong \mathbb{Z}_{p^k},$ every non-trivial subgroup $H$ contains or it is contained in the rest of subgroups. In other words, in this particular case, the subgroup lattice of $G$ is a chain. However, for me it is enough if exists some $H$ like this.
I wonder if it is possible to derive some result characterizing groups having a subgroup $H$ as above in the general case of finite groups (or at least abelian ones). Any help is very welcome.
Thank you for your time in advance!
 A: Suppose that $G$ is finite, and let $H$ be a nontrivial proper subgroup such that $H' \leq H$ or $H \leq H'$ for all subgroups $H' \leq G$.
I will leave the details to you, but here is a start:

Take some element $x \in G$ and $x \not\in H$. Then $\langle x \rangle$ must contain $H$, so...?

In the end you should be able to see that $G$ is a $p$-group with a unique subgroup of order $p$. By a classical theorem, such a group is cyclic or generalized quaternion.
A: I am going to write down the details of the solution proposed by spin. Thanks again.
Let me begin by giving a name to the property I am studying.
Definition: A group $G$ is said to be chain-reducible if there is a non-trivial subgroup $H$ $(H≠\{e\},G)$ such that $H′ \leq H$ or $H\leq H′$ for every subgroup $H′\neq H$. Otherwise, we say that $G$ is chain-irreducible.
Lemma 1: Let $G$ be a finite chain-reducible group. Then there is some prime $p$ such that $G$ is a $p$-group with a unique subgroup of order $p.$
Proof: As $G$ is chain-reducible there is a non-trivial subgroup $H$ $(H≠\{e\},G)$ such that $H′ \leq H$ or $H\leq H′$ for every subgroup $H′\neq H$. Suppose that $G$ is not a $p$-group, and the prime factorization of the order of $G$ is given by $\vert G \vert = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ with $r>1.$ Using Sylow theorems we know that there are Sylow $p$-subgroups $H_1, H_2$ of orders $p_1^{k_1}$ and $p_2^{k_2}$ respectively. If $H \leq H_1$ and $H \leq H_2$ then by Lagrange theorem, $\vert H \vert $ divides  $\vert H_1 \vert =p_1^{k_1} $ and $\vert H_2 \vert =p_2^{k_2} $ so $H$ is trivial (a contradiction). The case $H_1 \leq H \leq H_2$ (and $H_2 \leq H \leq H_1$) leads again to a contradiction. Therefore, It should be $H_1 \leq H$ and $H_2 \leq H.$ Now consider other Sylow $p$-subgroup $H_i$ of order $p_i^{k_i}$ with $i>2.$ If $H \leq H_i,$ as $H_1 \leq H$ then $p_1^{k_1}$ divides $p_i^{k_i}$ (contradiction) so $H_i \leq H$. If we repeat this procedure for every $i>1$ we find that $p_i^{k_i}$ divides $\vert G \vert$ for every $i,$ thus $H=G$ (a contradiction). So $r=1$ and $G$ must be a $p$-group with $\vert G \vert = p^k.$
Now let us prove that there is just one subgroup of order $p$. Consider $x\notin H,$ then $H \leq \langle x \rangle,$ so $H$ is a cyclic group. As $H$ is a finite cyclic $p$-group then $H$ is isomorphic to $\mathbb{Z}_{p^s}$ having just one subgroup of order $p.$ $\square$
Now I am going to use some classical results:
Theorem 1: Let $G$ be a finite $p$-group such that every abelian subgroup is cyclic. Then $G$ is either cyclic or a generalized quaternion group $Q_{4n}$ (or order $4n$) defined as:
$$Q_{4n}=< x,y \ \vert \ x^{2n}=y^4=1, x^n=y^2, y^{-1}xy=x^{-1} >.$$
(see https://en.wikipedia.org/wiki/Quaternion_group)
Theorem 2: A $p$-group is cyclic if and only if it is an abelian group having a unique subgroup of order $p.$
(see math.stackexchange.com/questions/4101103)
Finally we can characterize chain-reducible groups.
Theorem 3: If a finite group $G$ is chain-reducible then $G$ is a cyclic $p$-group (isomorphic to $\mathbb{Z}_{p^s}$) or the generalized quaternion group $Q_{4n}.$
Proof: Using Lemma 1 we know that $G$ is a $p$-group with just one element of order $p.$ Every abelian subgroup $F \leq G$ has order $p^s$ and has a unique $p$-subgroup of order $p$ (because there is just one in $G$). So we can use Theorem 2 to assure that every abelian subgroup is cyclic. Finally Theorem 1 gives the result. $\square$
