Corollary 2

Suppose that the group K operates freely on Som(G). There then exists a unique group homomorphism $\pi : (G/K)_{\gamma(o)}\rightarrow$ K whose kernel contains the image of $\gamma_o$ and such that $\pi (\lambda ([k])) =k$ for every $k\in K$. Moreover, $G_o\longrightarrow^{\gamma_o}(G/K)_{\gamma(o)}-\rightarrow^{\pi}$ K is an extension of K by $G_o$.

If such a group homomorphism $\pi$ exists, the group homomorphism $\pi \circ \lambda$ is necessarily equal to the unique group homomorphism $p$ from $G_o*F(K)$ into K such that $p(f) =e$ for $f\in G_o$ and $p([k]) =k$. It is immediate to verify that the elements of $G_o*F(K)$ defined by formulas ($R_2$) and ($R''_3$) belong to the kernel of $p$. By assumption, there is no element of type ($R'_3$). Thus, the kernel of the morphism $\lambda$ contains that of $p$. Consequently, there exists a unique group homomorphism $\pi : (G/K)_{\gamma(o)}\rightarrow K$ such that $\pi \circ \lambda =p$.

It is evident that the homomorphism $\pi$ is surjective. To prove that the homomorphism $\gamma_o$ is injective and that its image is exactly the kernel of $\pi$, let us remark that the homomorphism $\lambda ': M\rightarrow (G/K)_{\gamma(o)}$ (II, p. 213, remark 1) is an isomorphism, since it has been assumed that K operates freely on Som(G). The composite homomorphism $(\lambda ')^{-1}\circ \gamma_o$ from $G_o$ into M is given by $f\mapsto (e, f)$, whereas the homomorphism $\pi \circ \lambda ': M\rightarrow K$ maps $(k, f)$ to $k$. The corollary follows.