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Problem A. 382. (October 2005)

A. 382. S and T are disjoint sets, * is a binary operation on the elements of S and o is a binary operation on the elements of T. (That is, if a,b\inS and c,d\inT, then a*b\inS and cod\inT). Each operation is associative. In other words, (S,*) and (T,o) are semigroups. It is also given that for every t\inT there are elements u,v\inT, such that uot=tov=t. Let f\colon S\to T denote an arbitrary mapping. Define the operation \otimes on the set S\cupT as follows:


a\otimes b = \cases{
a * b & \quad \text{if}\quad a,b\in S, 
\\
f(a) \circ b & \quad \text{if}\quad a\in S,\ b \in T, \\
a \circ f(b) & \quad \text{if}\quad a\in T,\ b\in S, \\
a \circ b & \quad \text{if}\quad a,b\in T.
}

Show that the operation \otimes is associative if and only if f is a homomorphism, that is, f(a*b)=f(a)of(b) for all a,b\inS.

Czech competition problem

(5 pont)

Deadline expired on 15 November 2005.


Solution. (a) If f is a homomorphism then \otimes is trivially associative.

(b) Suppose that \otimes is associative and let a,b be two arbitrary elements of S. By the conditions there exist u,v\inT such that uof(a)=f(a) and f(a*b)ov=f(a*b). Then

f(a*b)=f(a*b)\circ v=(a\otimes b)\otimes v = a \otimes (b\otimes v) =
f(a)\circ(f(b)\circ v) = f(a)\circ f(b).


Statistics:

15 students sent a solution.
5 points:Erdélyi Márton, Estélyi István, Fischer Richárd, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, Kónya 495 Gábor, Korándi Dániel, Nagy 224 Csaba, Paulin Roland.
4 points:Kisfaludi-Bak Sándor, Tomon István.
2 points:2 students.
1 point:1 student.

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