Characterising finite groups $G$ such that every proper subgroup of $G$ has a proper supplement. Let $G$ be a finite group and $H$ be a proper subgroup of $G$. A proper subgroup $K$ of $G$ is said to be a supplement of $H$ in $G$ if $HK=G$. It is obvious that if the Frattini subgroup of $G$, $\Phi(G)$ is non-trivial, then $\Phi(G)$ and its non-trivial subgroups do not have any proper supplement in $G$.
My question is: Can we characterise finite groups $G$ with $|\Phi(G)|=1$ such that every proper subgroup of $G$ has a proper supplement?
 A: This is a partial answer, but the groups in question must be solvable. To see this, let $G$ a finite group with $\Phi(G)=1$ and such that every proper subgroup has a proper supplement. Choose $p$ the smallest prime dividing $|G|$. By Cauchy's Theorem we can find an $x \in G$ with $o(x)=p$. Put $H=\langle x \rangle$. If $G=H$ then $G$ is even cyclic whence solvable. So assume $H \lt G$. By the assumption, we can find a $K \lt G$ with $G=HK$. Since $K$ is proper we must have $H \cap K=1$ (if $H \cap K \gt 1$ then, since $H$ is of prime order, we would have $H \subseteq K$, so $G=K$ a contradiction). So $|G|=|HK|=\frac{|H||K|}{|H \cap K|}=|K| \cdot p$. Hence $|G:K|=p$, is the smallest prime dividing $|G|$, so $K$ must be normal. 
We will show that $K$ has the same properties as $G$ and by induction we then have $G$ is solvable by cyclic, whence solvable. Now, $K \lhd G$, so $\Phi(K) \subseteq \Phi(G)=1$. If $L \lt K$ is a proper subgroup of $K$, then certainly $L \lt G$, so it has a proper supplement $M$, that is $G=ML$. By applying Dedekind's Lemma we see that $K=K \cap ML=(K \cap M)L$. hence $K \cap M$ is a supplement to $L$. We finish by showing that $K \cap M$ is a proper subgroup of $K$: if not then $K \cap M=K$, or equivalently, $K \subseteq M$. But then $G=ML \subseteq MK=M$, contradicting the fact that $M$ is proper in $G$.
