Let G be a non-empty finite set, in which a multiplication ab is defined, which satisfies the following conditions.
( 1)(ab)c=a(bc)？
(2)ab=ac implies B = C.
(3)ac=bc implies a=b?
It is proved that G forms a group under this multiplication.
It is proved that 1) From (1) we know that the multiplication in G satisfies the associative law;
2) Let G be a non-empty finite set. Let the unary A in G be selected and the set aG={ax|x belongs to G}. Obviously, G contains aG, | ag |≤| g |. According to (2), f(x)=ax is the injectivity from G to aG, so | g |≤| ag |, so | ag |.
Take any unary A in G, let AX = A and XA = A be UA and VA respectively, that is, there are AUA = A and VAA = A. It is proved that ua=va is a unary G, because for any unary X in G, there is any unary U, V has Au = X and VA = X, so
xua=(va)ua=v(aua)=？ va=x
vax=va(au)=(vaa)u=au=x
According to the arbitrariness of X, UA and VA are left and right unitary elements of G, so va=vaua=ua, unitary elements exist;
3) Let the unitary element of G be E, then there is any element U in G, and V has xu=vx=e, that is, any X in G has left and right inverses, and v=ve=v(xu)=(vx)u=eu=u, that is, the left and right inverses are equal, and u=v is the inverse of X;
According to 1)2)3), G forms a group under this multiplication.
It is necessary for this problem to require that G is a finite set. If g is an infinite set, the conclusion of this problem is not valid. Considering the set of positive integers, the condition (1) (2) and (3) in the problem is obviously satisfied for ordinary multiplication on this set, but the set of positive integers does not form a group for ordinary multiplication.