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By Akira Takayama

An available advent to the analytical beginning of economics

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K,. different from zero. If here hiP', then m > 0. , in place and carries all other generators a. ', since goes into 1 under this ill r endomorphism, because in f the sum of the exponents for every generator is 0. a! sH, i= 1, 2, ... , m, 1 and therefore f EH, that il, fe(HOF')=K. (8) § 37. ' SUBGROUPS. IDENTICAL RELATIONS 47 Let n be the smallest positive exponent with which any generator a. l: F'. But since there exists an endomorphism of F that carries a. F". By (8) every exponent k1, k,, ••• , k..

L 3 is an arbitrary element of H then, as has been shown at the beginning of the section, the equation .. X 1 ••• XI =" , I = t1 == (where i~ if and only if a 11 a,) is an identical relation in F I H. By the isomorphism (10) it is also an identical relation in F /K; therefore the substitution of arbitrary elements of F in its left-hand side leads to an element of K. 1 we obtain h, which is therefore in the subgroup K, Thus H s; K. Similarly K s; H, and soH= K, and this is what we had to prove. Now we shall indicate the connection between the problem of a survey of all fully invariant subgroups of a free group F of arbitrary nMk and the corresponding problem for the case of a free group W of countable nMk.

LEMMA 1. Let A and B be isomorphic subgroups of a group G and let t:p be an isomorphic mapping of A onto B. Then G can be embedded in a group H containing an element h such that the transformation of A by h induces the mapping t:p, h- 1 ah= aqJ for all a of A. We consider the free products where { u} and { v} are infinite cyclic groups. J )v- 1 for all a of A. We can therefore construct the free product H of K and L with an amalgamated subgroup (see§ 35) by amalgamating U and V in accordance with the isomorphism 'I'.

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