Endomorphism of Brandt Semigroup $B_n(G)$, where $G$ is a finite group

Question

I want to show that $End_0 (B_n(G)) = \cup\phi_{\sigma,g} \cup C_{I(B_n(G))}$, where $\phi_{\sigma,g} : B_n(G) \rightarrow B_n(G) $ is an endomorphism is defined by $(i,a,j)\phi_{\sigma,g} = (i\sigma , ag , j\sigma)$ and $\sigma \in S_n$ and $g \in End(G)$, $C_X$ is the set of all constant map on $X$ and $I(B_n(G))$ is the set of all idempotents in $B_n(G)$.

I have proved that $\phi_{\sigma,g}$ and constant maps are endomorphism, so $ \cup\phi_{\sigma,g} \cup C_{I(B_n(G))} \subseteq End_0 (B_n(G))$

we have proved the converse part:

Let $\theta \in End_0(B_n(G))-C_{I(B_n(G))}$, then define $\theta_i : [n] \times G \times [n] : \rightarrow [n]$, where i=1,2 such that $(i,a,j) \theta = ((i,a,j) \theta_1 , b , (i,a,j) \theta_2)$, for some $b \in G$, where $[n] = \{ 1,2, \cdots , n\}$

Some work I have done , we get for any $k \in [n]$ $$ (l,a,k) \theta_2 = (k,e,j) \theta_1$$ $$(l,a,k) \theta_1 = (l,a,j) \theta_1$$ $$(l,a,j) \theta_2 = (k,e,j) \theta_2$$

we conclude that $\theta_1$ depends only first cordinates and $\theta_2$ depends only third cordinate.So we define $\sigma$ on $[n]$ such that $$ i \sigma = (i,e,k)\theta_1 = (k,a,i)\theta_2$$

which is a bijective map. Then $(i,a,j) \theta = ((i,a,j) \theta_1 , b , (i,a,j) \theta_2) = (i \sigma , b , j \sigma)$.

Similarly I want to define an endomorphism $f$ on $G$ such that $(i,a,j)\theta = (i\sigma , af , j\sigma)$

If we define $f : G \rightarrow G$ is such that $(i,a,j )\theta = (i \sigma , b , j \sigma)$, then $af = b$. Then $f$ is well defined if $(i,a,j )\theta = (i \sigma , b , j \sigma)$, where $b \in G$ is fixed and $\forall \ \ i,j \in [n]$.

I am unable to show that $(i,a,j )\theta = (i \sigma , b , j \sigma)$, where $b \in G$ is fixed and $\forall \ \ i,j \in [n]$.

Any help would be appreciated. Thank you

Answer 1

Note that $(i,a,j)=(i,1,1)(1,a,1)(1,1,j)$ and so a homomorphism is determined by what it does to the elements $(i,1,1)$ and $(1,1,j)$ and $(1,a,1)$ with $a\in G$. Moreover, since $B_n(G)$ is an inverse semigroup and $(i,1,1)$ is the inverse of $(1,1,i)$, in fact an endomorphism is determined by what it does to elements of the form $(1,1,i)$ and $(1,a,1)$ with $a\in G$ and $i,j$ in $\{1,\ldots,n\}$. Namely if $(1,1,i)$ maps to $(j,b,k)$ then $(i,1,1)$ maps to $(k,b^{-1},j)$ (or both map to $0$).

If two elements $(1,1,i)$ and $(1,1,j)$ with $j\neq i$ are collapsed, then the whole semigroup is collapsed since $(k,a,\ell)=(k,1,1)(1,a,1)(1,1,i)(i,1,\ell)$ and $0=(1,1,j)(i,1,\ell)$. So you get a constant map to an idempotent. Thus we may assume that no such collapse occurs.

Since the elements of the form $(1,1,i)$ all generate the same right ideal, the same is true for their images under an endomorphism. So $(1,1,i)$ maps to $(k,i\tau,i\sigma)$ where $\sigma$ is a permutation, $\tau\colon \{1,\ldots,n\}\to G$ is a mapping and $k$ is fixed. Since $(1,1,1)=(1,1,1)(1,1,1)$ it follows that $k=\sigma(1)$. Thus $(1,1,i)$ maps to $(1\sigma,i\tau,i\sigma)$.

Since $(1,a,1)=(1,1,1)(1,a,1)(1,1,1)$ it follows that $(1,a,1)$ maps to $(1\sigma,a\phi,1\sigma)$ where one checks directly that $\phi$ is an endomorphism of $G$.

So $(i,a,j)=(i,1,1)(1,a,1)(1,1,j)$ maps to $(i\sigma,(i\tau^{-1})a\phi (j\tau),j\sigma)$.

Conversely, any such $\sigma,\tau,\phi$ are easily checked to give an endomorphism. You were basically missing $\tau$.