Homework 04

Read: Finish 2.3 and skim through Appendix C: Definitions in Mathematics

Turn in: 2.35(e), 2.39, 2.41, 2.43, 2.44, 2.49

Remarks

• A comment about 2.39. This theorem (and many others) can be proved multiplying both sides of an equation by the same thing. However, since the operation may not be commutative, if you multiply one side of the equation by something on the left, then you need to do the same on the other side too. For example, if you know that $gx=gy$, then for any $h\in G$ it is also true that $hgx=hgy$; however, $hgx=gyh$ might not be true…do you see the difference? Now, to solve 2.39, choose something clever for $h$, so that you arrive at $x=y$.

• On 2.41, you might consider starting as follows.

  Let $G$ be a group, and let $g\in G$. Assume that there exists $g_1,g_2 \in G$ such that $gg_1 = g_1g = e$ and $gg_2 = g_2 g = e$. We will show that $g_1=g_2$. (Now keep going. Remeber that you can use previous theorems if you want.)

• Please make sure to look closely at 2.47 and 2.48 (and try to do them too). We’ll talk through them together in class next time. To clarify 2.48: the goal is to simply rewrite Theorem 2.47 using “additive notation.” See Remark 2.46 in the book. The point is that for an abstract group, $gh$ and $g+h$ mean the same thing (as do $g^n$ and $ng$); we are just representing the binary operation multiplicatively in the first case and additively in the second. We will never mix multiplicative and additive notation in the same problem, but some problems will use one and some the other. Additive notation is typically only used with abelian groups.