Read: Beginning of Section 2.3, and Appendix A: Elements of Style for Proofs
Turn in: (Solutions to or your best efforts on) 2.31, 2.35(b,c), 2.36, 2.37
- Although I’m not asking you to prove 2.29, please read it over, and know that you can use it in the future. That theorem is something proved in our Math 108 course.
- About the definition of a group: Axiom 0 refers to “closure” which was defined in the discussion following Definition 2.15. It is only for emphasis since if $*$ is a binary operation on $G$, then $G$ must be closed under $*$. In short, if you already know $*$ is a binary operation on $G$ then you do not need to verify Axiom 0 again.
- About 2.35: please make sure to explain why each of the group axioms is true. Remember to use our previous work, so in each case of 2.35 you already know that composition is a binary operation on the set. Thus, it only remains to check Axioms 1,2,3. Also, since the operation is function composition, you can use 2.29 for associativity. So really the work is to identify an identity element (satisfying Axiom 2) and to identify an inverse $g’$ of each element $g$ (satisfying Axiom 3).
- About notation: in this book, $\mathbb{Z}$ refers to the integers and $\mathbb{N}$ refers to the positive integers (i.e. $\mathbb{N} = \{1,2,3,\ldots\}$). In 2.36(g), the set $\mathbb{Q}\backslash \{0\}$ is the set of all rational numbers excluding 0. In general, if $A$ and $B$ are sets, then $A \backslash B$ is the set set of all elements of $A$ that are not also in $B$.
- About 2.37: a good way to show that there is only one element with a certain property is to assume you have two elements with that property and then show that they are the same. So here is one way you might start your proof of 2.37.
Let $G$ be a group. Assume that $e_1$ and $e_2$ are identity elements. We will show that $e_1=e_2$. (Now keep going using that both $e_1$ and $e_2$ satisfy the equation given in Axiom 2 of a group.)
