On normal $p$-complements This is question 5E.3. of Isaacs's Finite Group Theory: 

Suppose every two generator subgroup of a finite group has a normal $p$-complement. Show that $G$ has a normal $p$-complement.

Of course there's a hint there: consider the subgroup generated by $x$ and $y$, where $x$ is in some $p$-group $P$ and $y \in {N_G(P)}$ has order not divisible by $p$.
Any hints on how to use the hint would be immensely appreciated!
 A: Let $X$ be a $p$-subgroup. We are going to show that $N_G(X)/C_G(X)$ is a $p$-group and then we can apply Frobenius' Theorem 5.26 to be found in Isaacs' Finite Group Theory. If $X$ is trivial there is nothing to prove so let $x \in X$, be a $p$-element. If $N_G(X)$ is a $p$-group, then certainly the quotient $N_G(X)/C_G(X)$ is, and we are done. So we can find a $y \in N_G(X)$, with $y$ a $p'$-element. The assumption assures that $K= \langle x,y \rangle$ has a normal $p$-complement, say $K=PL$, where $P \in Syl_p(K)$ is chosen such that $x \in P$, $L \unlhd K$, $L \cap P=1$. Observe that $y \in L$, since $K/L$ is a $p$-group and $y$ is a $p'$-element. Note that $L$ is a $p'$-subgroup of $K$, which implies that $X \cap L=1$.Now let us have a look at $[x,y]=x^{-1}y^{-1}xy$. Since $y$ normalizes $X$, we see that $x^{-1}(y^{-1}xy) \in X$. Since $L$ is normal in $K$, we also see that $(x^{-1}y^{-1}x)y \in L$. So the commutator $x^{-1}y^{-1}xy \in X \cap L=1$. Hence $x$ and $y$ are commuting, which makes $K$ even abelian. What we have proved is that in $N_G(X)$ every $p'$-element commutes with $X$. Which means that if $Q \in Syl_q(N_G(X))$, with $q \neq p$, we must have $Q \subseteq C_{N_G(X)}(X)=C_G(X) \cap N_G(X)=C_G(X)$. This implies that $|N_G(X):C_G(X)|$ must be a $p$-power, and we are done.
