Inverse Binary Operations

——————————————————————————————————————————Identity: Let e ∈ G be the identity element of (G, •). Then, for any a ∈ G we havee • a = ae ∗ a −1 = ae ∗ a = a,where we have used the fact that every element in (G, ∗) is its own inverse. As e ∗ a = a musthold for all a ∈ G, e must also be the identity element of (G, ∗). Note that the other condition(a • e = a) is now automatically satisfied.——————————————————————————————————————————Inverse: Fix a ∈ G. Let b ∈ G be the inverse of a with respect to •. Then we havea • b = ea ∗ b −1 = ea ∗ b = e,where we have used the fact that every element in (G, ∗) is its own inverse. By the uniquenessof inverses in (G, ∗) we must have b = a −1 . Note that the other condition (b • a = e) is nowautomatically satisfied.——————————————————————————————————————————So, the necessary condition for (G, •) to be a group is that every non-identity element of G hasorder two with respect to ∗ (we did not need to impose any further assumptions to guarantee theexistence of an identity and inverses). Not only is this condition necessary, it is also sufficient (thethree axioms are satisfied by construction).If (G, ∗) is a group such that every non-identity element has order 2, then (G, •) is a group.The identity element with respect to • is the identity element with respect to ∗, and the inverseof a with respect to • is the inverse of a with respect to ∗. Given that these two groups have thesame identity and inverses, are they really different groups? No! Note thata • b = a ∗ b −1 = a ∗ b,where we have used the fact that each element is its own inverse. Therefore • = ∗; that is, (G, ∗)and (G, •) are equal as groups – not just isomorphic, actually equal! It is worth noting that thisrealisation could have been made after enforcing associativity alone, in which case the fact thatthe two groups have the same identity and inverses would be obvious.