1. Let g be an element of a group (G,*) such that for some one element xis in G,
ID: 2940718 • Letter: 1
Question
1. Let g be an element of a group (G,*) such that for some one element xis in G, x*g=x. Show that g=e. 2. Let (G,*) be a group. Show that (G,*) is abelian if and only if (x*y)-1 = x-1 * y-1 for all x,y in G. Please help me to try and figure out how to finish these proofs.1. Let g be an element of a group (G,*) such that for some one element xis in G, x*g=x. Show that g=e. 2. Let (G,*) be a group. Show that (G,*) is abelian if and only if (x*y)-1 = x-1 * y-1 for all x,y in G. Please help me to try and figure out how to finish these proofs.
Explanation / Answer
1. By contrapositive, suppose g not= e then x*g not= x*e=x ==> x*g not=x. 2. Suppose G is abelian. Then for all x,y in G x*y=y*x. Essentially this means we can flip around the operation "*". So in any group (x*y)^-1=y^-1 * x^-1, but since G is abelian we can flip around the right side of the equation to get: (x*y)^-1=x^-1 * y^-1 Suppose for all x,y in G (x*y)^-1=x^-1 * y^-1. Note that the following equation still holds: (x*y)^-1=y^-1 * x^-1. Thus they must be equal: x^-1 * y^-1 = y^-1 * x^-1. Thus G is abelian. P.S. don t let the negative exponent bother you. its true for all g in G because all g have an inverse.
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