TITLE: Is a finite group with any subgroup admitting a unique complement, cyclic?
QUESTION [5 upvotes]: Let $G$ be a finite group such that any subgroup $H \le G$ admits a unique complement $K$.  
Question: Is $G$ cyclic ?

REPLY [6 votes]: This result admits a proof at the subgroups lattice level. 
Let $\mathcal{L}(G)$ be the subgroups lattice of $G$, then we see that $\mathcal{L}(G)$ does not admit sublattice equivalent to the following diamond or pentagon lattices

But there is the following result (see theorem 101 p 109 in this book): 

Theorem: A lattice is distributive iff it has no sublattice equivalent to the diamond lattice or the pentagon lattice.

It follows that the lattice $\mathcal{L}(G)$ is distributive.  
The result follows by a result of Oystein Ore of 1938 (theorem 4 p 267 in this paper): 

Theorem: A finite group is cyclic iff its subgroups lattice is distributive.

Remark: we observe that $\vert G \vert$ is square-free.
Remark: as a corollary of the proof, we see that the more generalized statement for which we replace "complement" by "lattice-complement" (i.e. $\langle H,K \rangle  =G $ and $H \cap K = 1$) is also true.