3
$\begingroup$

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 $(S\times T,\circ)$?

$\endgroup$

closed as off-topic by Stefan Kohl, Benjamin Steinberg, YCor, Derek Holt, S. Carnahan Jun 3 '14 at 22:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Derek Holt, S. Carnahan
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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 '14 at 17:33
  • 3
    $\begingroup$ This has also been posted to math.stackexchange.com/questions/819509 $\endgroup$ – Derek Holt Jun 3 '14 at 17:35
  • $\begingroup$ Yes I posted there. Is it illegal?If yes I'll remove it! $\endgroup$ – David Jun 3 '14 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. $\endgroup$ – Benjamin Steinberg Jun 3 '14 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 '14 at 18:34