We are in the contest of the classification of all groups of order $2^3$.
We know that a group $G$ with a unique maximal subgroup is necessarely cyclic.
1)Then my teacher said that the dual proposition is "almost" true: a group $G$ with a unique minimal subgroup is a $p$-group and if $p\neq2$ it's also cyclic. Is this true? After this he said that there is a unique counterexample to this (he spoke about $Q_8$, but it seems it respects the dual proposition: $\langle-1\rangle$ is its unique minimal subgroup). So what is this group that makes exception?
2)Then he said: suppose $G$ s.t. $H\unlhd G\;\;\forall\;\;H\le G$, then either $G$ is abelian, or $G$ admits $Q_8$ as a direct factor (i.e. $\exists A\le G$ s.t.$G=Q_8\times A$): why? Finally, from this it follows that the only two groups of order $2^3$ are $D_4$ or $Q_8$. WHY? I'm struggling on this. I found on the web some useful pdf that explain why the only two groups of order $2^3$ are these two... but it was made only with computations, and I'm really interested in understand this in the way my teacher did: it seems smarter, mathematically better.
If someone can help me I would be really grateful. I'm aware my questions could seem a little bit confusing, but this is the best can I did with my notes. Thank you all.
Most of the claims are close to correct, but are missing hypotheses. Here are the correct claims with proofs or references.
Claim 1: If $G$ is a finite group with a unique minimal subgroup, then $G$ is a $p$-group.
Proof: If $|G|$ were divisible by two distinct primes $p,q$, then Cauchy's theorem would guarantee the existence of two subgroup $P,Q$ of order $p,q$. Since a prime order subgroup is minimal, this gives two minimal subgroups a contradiction. Hence $|G|$ is divisible by at most one prime, so $G$ is a $p$-group. $\square$
Counterexample 1: There are infinite groups $G$ with unique minimal subgroups that are not $p$-groups.
Proof: $G=\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is one such group. A minimal subgroup always has prime order, and the subgroups of finite order of an abelian group all live inside the torsion subgroup, in this case $0 \times \mathbb{Z}/2\mathbb{Z}$, which clearly has a unique minimal subgroup. $G$ itself is not a $p$-group, since it has elements of infinite order. $\square$
Claim 2: If $G$ is a finite group of odd order with a unique minimal subgroup, then $G$ is cyclic of prime power order.
Proof: See Gorenstein's Finite Groups Theorem 5.4.10 on page 199, or M. Hall's Theory of Groups Theorem 12.5.2 on page 189. $\square$
Counterexample 2: There are infinitely many non-isomorphic non-cyclic finite groups of order a power of 2 that have a unique minimal subgroup.
Proof: The generalized quaternion subgroups of order $2^{n+2}$ for $n\geq 1$ have a unique subgroup of order 2, so a unique minimal subgroup. These are in fact the only non-cyclic finite groups with a unique minimal subgroup, again by Theorem 5.4.10 on page 199 or Theorem 12.5.2 on page 189. $\square$
Claim 3: If $G$ is a non-abelian group in which every subgroup is normal, then $G$ is a direct product of $Q_8$ and some subgroup.
Proof: These are called Hamiltonian groups. These all have the structure $Q_8 \times A \times E$ where $A$ is an abelian group in which every element is of finite odd order, and $E$ is an abelian group in which every element has order 1 or 2. This is proven as Theorem 12.5.4 on pages 190–192 of Hall's Theory of Groups. $\square$