Remark 1

One can give another description, sometimes more convenient, of the group $(G/K)_{\gamma(o)}$. For this, let $M = K\times G_o$ and define a law of composition in M by the formula

for $k,h\in K$ and $a,b\in G_o$. One verifies that this law of composition is associative, that $(e, c^{-1}_e)$ is an identity element, and that the element $(k^{-1}, c^{-1_1}_{k^-}(k^{-1})^*(c_ka)^{-1}c_e)$ is the inverse of $(k, a)$. It therefore endows M with a group structure. Moreover, the mapping $\lambda ': M\rightarrow (G/K)_{\gamma(o)}$ defined by $(k, a)\mapsto \lambda ([k]a)$ is a group homomorphism. Let $\alpha '$ be the unique group morphism of $G_o*F(K)$ into M such that $\alpha '(f) =$ $(e, f)$ if $f\in G_o$ and $\alpha '([k]) = (k, e_o)$ if $k\in K$; one has $\lambda '\circ \alpha '=\lambda$.

The relations ($R_2$) and ($R''_3$) show that every element of $(G/K)_{\gamma(o)}$ is the image under the homomorphism $\lambda$ of an element of $G_o*F(K)$ of the form $[k]f$, with $f\in G_o$ and $k\in K$. Hence the homomorphism $\lambda '$ is surjective. One verifies moreover that the image under $\alpha '$ of an element of $G_o*$ F(K) of the form $(R_2)$ or $(R''_3)$ is equal to the identity element. Consequently, the kernel of the homomorphism $\lambda '$ is the smallest normal subgroup of M containing the images under $\alpha '$ of the elements of $G_o*F(K)$ of the form ($R'_3$).