Read: Continue with 4.1.
Turn in: 4.12(i,k), 4.19, 4.24
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4.19 is a challenging one. There are several things to show: assuming $n$ is the smallest positive integer such that $g^n=e$, you want to prove (1) that \(\langle g \rangle = \{e,g,g^2,\ldots,g^{n-1}\}\) and (2) that the elements in the list \(e,g,g^2,\ldots,g^{n-1}\) are all distinct.
For (1), you are proving equality of sets. Note that \(\{e,g,g^2,\ldots,g^{n-1}\}\subseteq \langle g \rangle\) follows quickly from the definition of \(\langle g \rangle\). To show \(\langle g \rangle \subseteq \{e,g,g^2,\ldots,g^{n-1}\}\), you want to prove that for any \(m\in \mathbb{Z}\), $g^m$ is equal to one of \(e,g,g^2,\ldots,g^{n-1}\). The Division Algorithm (Theorem 4.16) is very helpful: write $m = nq+r$ with $0\le r < n$ and then $g^m = g^{nq +r}$. Keep going (using that $g^n=e$)…
For (2), assume (towards a contradiction) that \(g^i = g^j\) for some $i,j$ with \(0\le i< j < n\). Try to manipulate things to contradict that $n$ is the smallest positive integer such that $g^n=e$.
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Note that 4.24 is an “if and only if” statement, so there are two things to show. For the “backward” direction, you will assume that $n$ divides $i-j$ (meaning that \(i-j = nk\) for some \(k\in \mathbb{Z}\)); what does this imply about $g^{i-j}$? For the “forward” direction, you assume $g^i = g^j$ and want to prove that $n$ divides $i-j$. One strategy for proving that $n$ divides $i-j$ is to use the division algorithm to write $i-j = qn +r$ for some $0\le r < n $ and then work to prove that $r = 0$.
Extra practice: 4.14, 4.20–4.23