1. The problem statement, all variables and given/known data If G is a group and a ∈ G, let π: G--> G be the function defined as π(g) = ag, for all g ∈G. a) Show that π is a bijection b) Show that if π is an isomorphism, then a is the identity element of G. 2. Relevant equations I think to show that pi is a bijection we have to show that it is surjective and injective. To show that it is an isomorphism we have to show that π(xy) = π(x)π(y). 3. The attempt at a solution First, I do part a and show that it is one to one: Suppose π(g) = π(g') then ag = ag' Since a ∈G we know a^(-1) exists in G. a^(-1)ag = a^(-1)ag' g = g' and so π is one to one. I am not sure how to prove that this is onto... any hints here would help. For onto can I just say Let x,y ∈ G and x = a^(-1)y since a is a part of the group we know that a^(-1) exists. And thus it follows that π(x) = y To show part b) if x,y ∈G, then we observe π(xy) = axy π(x)π(y) = axay Here, a must equal the identity element of G for this function to be an isomorphism.