Problem from an old prelim exam:
If $G$ is a group and $H$ is a subgroup, then a normal complement to $H$ in $G$ is a normal subgroup $N\trianglelefteq G$ such that $N\cap H = \{1\}$ and $N H = G$. Show that if $G$ is a finite group in which every subgroup has a normal complement, then $G$ is nilpotent.
My first thought is to use the theorem (6.3.2 in Dummit & Foote) that $G$ is nilpotent iff $H<N_G(H)$ for every proper subgroup $H<G$.
Thus, let $H<G$ be proper; it suffices to find $g\in G\setminus H$ such that $H^g=H$. The assumption gives $N\trianglelefteq G$ such that $G\cong N\ltimes H$, so, in order that $H^g=H$, we need $g=(n,k)$ such that for every $h\in H$ we have
$$\begin{align} g (1, h) g^{-1}&=(n, k) (1, h) (n, k)^{-1}\\ &=(n (k\cdot h\cdot k^{-1}\cdot n^{-1}),khk^{-1})\\ &\in H. \end{align}$$
That is, we need nontrivial $n$ such that $n=h^k\cdot n$ for every $h$. Clearly, such an $n$ need not exist for every choice of $N,H$ (for example, $N\ltimes \text{Aut}(N)$), so I guess we need to apply the normal complement assumption to a subgroup other than $H$ itself. But which one? Or should we use a different approach entirely?
The assumptions actually imply that $G$ must be abelian. One can prove this without the assumption that $G$ is finite.
Let $x\in G$ and let $N_x$ be a normal complement to the cyclic subgroup $\langle x\rangle$. By the Second Isomorphism Theorem, $G/N_x\cong \langle x\rangle$. Since $\langle x\rangle$ is disjoint from $N_x$, the intersection $I:=\bigcap N_x$ of all $N_x$'s is disjoint from every cyclic subgroup of $G$, hence equals $\{1\}$. This yields $$G\cong G/I=G/\bigcap N_x \leq \prod (G/N_x)\cong \prod \langle x\rangle.$$ The cyclic groups $\langle x\rangle$ are abelian, and abelianness is inherited by products and subgroups, so $G$ is abelian. \\\