Let G be a group with identity e and product ab for any elements a and b of G.
Let ф: G⟶G be a map such that Ф(a sub1)ф (a sub2)ф(a sub3) = ф(b sub1) ф(b sub2) ф(b sub3) whenever,
(a1) (a2)(a3) = e=(b1) (b2) (b3) for any(not necessarily distinct) elements a1 ,a2 ,a3, b1, b2, b3 of G.
Prove: There exists an element a in G such that the map @: G⟶G defined by @(x) = a ф(x) for every element x in G, is a homomorphism.
Def: A homommorphism Ф from a group G to a group G is a mapping from G to G that preserves the group opperation. That is, Ф(ab) = Ф(a)Ф(b) for all a, b in G
Properties of elements under a homomorphism:
Ф carries the identity in G to the identity in G
Ф preserves inverses
*note that the Ф in this section is not the same as in the question...
The Attempt at a Solution
I really haven't got a clue on where to even begin to define "a" I am thinking that it needs to be triplet for example aea^(-1)...but I have no idea what to do or where to start...I am completely lost, can anyone give me a push in the right direction?