Let $\pi$ be a set of primes, let $P$ be a Hall $\pi$-subgroup of a finite $G$ and let $G'(\pi)$ be the inverse image of $O_{\pi'}(G/G')$ in $G$. Consider the subgroup $P^*=\langle[y,g]:y,y^g\in P, g\in G\rangle$ and let $\tau: G\to P/P'$ be the transfer map. Then $P/P'=(P^*/P')\tau(G)$.
Here $\pi'$ is the set of all primes not in $\pi$ and $O_{\pi'}(G/G')$ is the smallest Hall $\pi'$-subgroup of $G/G'$ (alternatively, the largest normal $\pi'$-subgroup).
Proof: Let $\overline P=P/P'$. Recall that $$\tau(x)\overline{P^*}=\overline{x}^{|G:P|}\overline{P^*}$$ for all $x\in P$. Since $P$ is a Hall $\pi$-subgroup, we get $$\langle\tau(x)\overline{P^*}\rangle=\langle\overline{x}\overline{P^*}\rangle.$$ Thus, $\overline{P\cap \ker \tau}\subseteq \overline{P^*}$ and so $$P\cap \ker \tau\subseteq P^* \text{ and }\overline{P}=\overline{P^*}\tau(G).$$
I'm having trouble understanding how the equality $\overline{P}=\overline{P^*}\tau(G)$ is concluded at the end. Of course, $\overline{P}\supseteq\overline{P^*}\tau(G)$ but I can't see why the other containment holds too. I tried showing their orders are equal but was not able to nor was I able to define an isomorphism between the two sets.
For $x\in P$ we have $\overline{x}\in\langle\overline{x}\overline{P*}\rangle=\langle\tau(x)\overline{P^*}\rangle\subseteq\tau(G)\overline{P^*}$.