Read: Finish Section 2.3. Consider also watching this Socratica video to revisit some of the things we’ve talked about so far: Group Definition (expanded) - Abstract Algebra | Socratica (11:14 min)
Turn in: 2.41, 2.43, 2.44, 2.49
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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.)
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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 if we write the operation multiplicatively, e.g. the group is \((G,*)\), then we’ll use the shorthand \(g^n=g*g*\cdots*g\) ($n$ times); whereas, if we write the operation additively, e.g. the group is \((G,+)\), then we’ll use the shorthand $ng=g+g+\cdots+g$ ($n$ times). 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.
Extra practice: 2.42