We have already seen (cf. Exercise 11, Chapter 1) that in a group ab = bc need n

Question

We have already seen (cf. Exercise 11, Chapter 1) that in a group ab = bc need not imply a = c. Suppose G is a group with the property that ab = bc does indeed imply that a = c for all a,b,c G G. Show that G must be abelian. Show, by explicit example, that "middle cancelation" is not necessarily valid in a group; that is, find a group G and elements a, b, c, d, x G such that axb = cxd, but ab cd. Prove that if a group G does satisfy the "middle cancelation" property above for all a. b, c, d, x G then G must be abelian.

Explanation / Answer

(first part) If ab = bc => a=c let x,y from G notice xyx = xyx using above formula, a= (xy) , b = x, c = (yx) Thus (xy)x = x(yx) => xy = yx => This is Abelian!

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