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Dec 21, 2016 at 18:31 vote accept user120386
Dec 21, 2016 at 18:17 comment added user120386 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 $\theta \ \ \in \ \ End(B_n(G))$.
Dec 21, 2016 at 18:13 comment added user120386 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 $(i,1,1)$.
Dec 21, 2016 at 17:35 answer added Benjamin Steinberg timeline score: 3
Dec 21, 2016 at 17:30 comment added Benjamin Steinberg You are missing one ingredient in your formulation, conjugation by a diagonal matrix.
Dec 21, 2016 at 17:24 comment added Benjamin Steinberg I will write up the proper answer and erase the comments.
Dec 21, 2016 at 16:21 comment added user120386 @ 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 $a \in \ \ G$.
Dec 21, 2016 at 10:20 review First posts
Dec 21, 2016 at 10:23
Dec 21, 2016 at 10:15 history asked user120386 CC BY-SA 3.0