Please Help, Abstract Algebra proof!?

Question

Let G be a group with the following property: whenever a, b, and c belong to G and ab = ca, then b = c. How can you prove that G is Abelian??? (hint: cross cancellation implies commutatively).

Answer 1

By definition:

Let (G, *) be a group. If for any a, b in G we have a*b = b*a, the group is abelian (or commutative).

You have a,b and c in G, the operation is (.)
You have defined a.b = c.a
Let's find b.a
b.a = a.c, a.c = c.a
Then, (G, .) is abelian.

Answer 2

Let a, b be any two elements in G and c=aba⁻¹. Then ca = aba⁻¹a = ab, so by the given property, c=b, thus ba = ca = ab. Since this holds for arbitrary a and b, G is abelian.

Answer 3

Given ab = ca, multiply on the right by a^-1 to get
aba^-1 = c
But b= c
so
a ba ^-1 = b
Multiply on the right by a
Then ab = ba for any a, b in G,
so G is abelian.

Answer 4

sub b=c in ab=ca then ab=ba so G is abelian

Answer 5

let x and y belongs to G then xyx=xyx and since associativity holds we have x(yx)=(xy)x.let a=x , b=yx ,c=xy:we have ab=ca=>b=c and yx =xy then G is abelian