Exercise 1) Let be a group and let be any element in . Prove that the inverse of is . I.e., show that “the inverse of the inverse is what you started with”. (Hint: this is completely trivial from the definition of a group. I.e., if we know that is the inverse of , then we simply take the properties that it has with respect to and note that they imply that also immediately has those properties with respect to . This is mainly an exercise in getting the flow of logic correct and precise.
Solution: As mentioned in the hint, this is mainly an exercise in getting the logic correct with arguments like these. Thus, even though some steps may seem trivial, they’re indeed important to understand fully. Let’s begin.
Pick any element in the group , and let the element denote a’s inverse (we know such an inverse exists because is a group, and inverses exist for all elements in a group by definition, and we know it’s unique by our work in this lesson). Now we want to find the inverse of since, after all, is an element of and therefore itself has an inverse. Well, we know that the inverse of must be an element such that , where is the identity element (we’ve often denoted this by “0” as well). But, we know that has the very same properties that does. Therefore , since we know that inverses are unique, and therefore is the inverse of , just as is the inverse of .
This shouldn’t be surprising since, for example, the inverse of is , and the inverse of is (when the group is the integers with addition). You should convince yourself that the groups we’ve discussed, and any that you may ever come up with, all have this property (as they must!).
On to lesson 24