Suppose that G is a group with the property that for every choice of elements in G, axb = cxd but ab = cd. Prove that G is Abelian. ?
("Middle cancellation" implies commutativity.)
- 3 weeks ago
If this property holds for all choices of a,b,c,d,x in G, then it must hold when b is the identity element of the group, which must exist since G is a group.
Letting b be the identity, this gives us that a = cd implies ax = cxd. Substituting in a = cd into the second statement, this gives us that cdx = cxd.
Now we can multiply both sides by the inverse element of c, which again must exist since G is a group. c^-1 * cdx = c^-1 * cxd.
Because G is a group, multiplication must be associative, so (c^-1 * c) * dx = (c^-1 * c)xd, which implies edx = exd or just dx = xd for all d and x in G.
The only additional requirement for a group to be Abelian is that its set operation is commutative for all elements in the group, so we're done.
I hope this helps, let me know if anything needs further clarification!