A semigroup $S$ acts on another semigroup $V$ (written additively for better readability, but could be non-commutative) on the left if $$ s(v_1 + v_2) = sv_1 + s v_2, \quad s(s')v = (ss')v $$ for $s,s' \in S, v,v_1,v_2\in V$. Similar a right action is given by the laws $(v_1 + v_2)s = v_1s + v_2s$ and $(vs)s' = v(ss')$.
Suppose we have three semigroups $S,T,V$ such that $S$ acts on $V$ on the left, and $T$ on the right, and such that both action are compatible in the sense that we have $$ (sv)t = s(vt) $$ for $s \in S, t \in T, v \in V$. Then we define the triple "semi-direct" product $(T,V,S)$ as the set of all triples $(t,v,s)$ with multiplication $$ (t,v,s)(t',v',v') = (tt', vt' + sv', ss'). $$
I have a question on the following proof:
Let $G$ be a group in $(T,V,S)$. There exists then an invariant (=normal) subgroup $H$ of $G$ such that $H$ is isomorphic with a group in $V$ while $G/H$ is isomorphic with a group in $T \times S$.
Proof. Let $(c,d,e)$ be the unit element of $G$. Then $c$ is an idempotent in $T$ and $e$ is an idempotent in $S$. Define $$ H = \{(c,v,e) \mid (c,v,e) \in G \}. $$ Clearly $H$ is a subgroup of $G$ and is the kernel of the morphism $\varphi : G \to T\times S$ given by $\varphi(s,v,t) = (s,t)$. Thus $G/H$ is isomorphic with a group in $T \times S$. To show that $H$ is isomorphic with a group in $V$ we define $\psi : H \to V$ by setting $$ \psi(c,v,e) = evc. $$ We have \begin{align*} \psi[(c,v,e)(c,v',e)] & = \psi(c,vc + ev', e) \\ & = e(vc + ev')c = evc + ev'c \\ & = \psi((c,v,e)) + \psi((c,v',e)) \end{align*} so that $\psi$ is a morphim. Since \begin{align*} (c,v,e) & = (c,d,e)(c,v,e)(c,d,e) \\ & = (c,dc + evc + ed,e) \end{align*} it follows that $\psi$ is injective. Thus $\psi$ maps $H$ isomorphically onto a group in $V$. $\square$
I do not understand the last step, in reasoning that $\psi$ is injective? It is just rewritten $(c,v,e)$ with the idenity $(c,d,e)$. As the image of $\psi$ is a group, the kernel in this case is the inverse image of the idenity, which is $edc$. Hence what we should compute is $$ \psi(c,v,e) = edv $$ which is equivalent with $evc = edv$. But I do not see how to go on from here, and what this has to do with the given computations? (Similar showing $evd = ev'd$ implies $v = v'$ I am not able to derive...)
Could anyone please explain what is happening here?
Remark: The statement is taken from page 145 of Samuel Eilenberg, Automata, Languages, and Machines, Vol. B.
Suppose that $\psi(c,v,e) = \psi(c,d,e)$, that is, $evc = edc$. Then $$ (c,v,e) = (c,dc+evc+ed,e) = (c,dc+edc+ed,e) = (c,d,e) $$