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Let $S,T$ be two semigroups. A function $f:S\to T$ is called multiplicative if for any $x,y\in S$ we have $f(xy)=f(x)f(y)$. If $T=S$ then $f:S\to S$ is called automorphism on $S$.

A function $f:S\to\Bbb{C}$ is semicharacter on the semigroup $S$, if it is multiplicative, where $\Bbb{C}$ is complex plane.

Let $S,T$ be two semigroups and $Aut(S)$ be the set of all automorphisms on $S$. Consider a multiplicative function $\phi:T\to Aut(S)$, and define a semidirect product on $S\times T$ as below

$$(s_1,t_1)\circ(s_2,t_2)=\Big(s_1([\phi(t_1)](s_2)),t_1t_2\Big)$$ It was proven $S\times T$ with this product is semigroup. Also $S\times T$ with the following product is semigroup $$(s_1,t_1).(s_2,t_2)=(s_1s_2,t_1t_2)$$ It was proven that for any semicharacter $f$ on $(S\times T,.)$ there exists two unique semicharacters $g$ on $S$ and $h$ on $T$ such that for all $x\in S,y\in T$ $$f(x,y)=g(x)h(y)$$.

Now the question is this that what we can say about semicharacters on $(S\times T,\circ)$?

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  • $\begingroup$ Do you not require automorphisms to be bijective? Also, $\phi(s_2)$ does not make sense, since the domain of $\phi$ is $T$. Do you mean $\phi(t_1)(s_2)$? $\endgroup$
    – Derek Holt
    Jun 3, 2014 at 17:33
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    $\begingroup$ This has also been posted to math.stackexchange.com/questions/819509 $\endgroup$
    – Derek Holt
    Jun 3, 2014 at 17:35
  • $\begingroup$ Yes I posted there. Is it illegal?If yes I'll remove it! $\endgroup$
    – David
    Jun 3, 2014 at 17:38
  • $\begingroup$ Characters of any semigroup factor through their commutative image. For characters on commutative semigroups look at Clifford and Preston's book. $\endgroup$ Jun 3, 2014 at 17:50
  • $\begingroup$ @David: usually after posting on MSE one waits a little time without satisfactory answer before posting to MO, and if so, it should be mentioned on both MO that is was posted on MSE with no answer, and on MSE that it's followed-up on MO. $\endgroup$
    – YCor
    Jun 3, 2014 at 18:34

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