J.-E .Pin : How both the set are equal

J.E Pin : I have deleted from math.stackexchange.com

on way for the proof is to show $T$ is isomorphic $ I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\}$ and other way is to show $T \subseteq I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\}$ and $ I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\} \subset eq $.

You mean to say that $B_{\lambda}^o(S)$ is a free congruence.

I know that if we define $g : G \rightarrow G$ such that $(1,a,1) \theta = (1\sigma , ga ,1\sigma)$. is an endomorphism on $G$ and $\theta \ \ \in \ \ End(B_n(G))$.

Since $(i,a,j) = (i,1,1) (1,a,1) (1,1,j)$ and $(1,a,1)$ maps to $(1 \sigma, ga , 1\sigma)$ and if $(i,1,1)$ maps to $(i \sigma , b , 1 \sigma)$ implies $(1,1,i)$ maps to $(1 \sigma , b^{-1} , i \sigma)$. But what is the image of $(i,1,1)$.

@ Benjamin : How we can write every element (i,a,j) is the product of the elements $(j,1,j)$ and $(1,a,1)$, where $j \in \ \ [n]$ and $a \in \ \ G$.