Read: Continue with 4.1.
Turn in: 4.25, 4.27 or 4.29 (you choose one), 4.36, 4.37
- Using 4.24 for 4.25 leads to a really short proof.
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If you choose to prove 4.27, you can just assume $n\ge 3$. That is, assume $G$ is a finite cyclic group of order $n$ with $n \ge 3$, and prove that \(G\cong R_n\). One way to start is to use 4.19 to write \(G = \{e,a,a^2,\ldots, a^{n-1}\}\) for some generator $a$ of $G$. Now, think about the elements of $R_n$, and use the similarities with your expression for $G$ to define a function \(\phi: G \rightarrow R_n\) that you suspect is an isomorphism. Then prove that it works.
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For 4.29 you want to construct an isomorphism between $G$ and $\mathbb{Z}$. I recommend defining it from $\mathbb{Z}$ to $G$. You know $G$ is cyclic, so \(G = \langle g \rangle = \{g^k \mid k \in \mathbb{Z}\}\). To define $\phi$, you need to decide how to complete this: \(\phi(n) = g^{??}\). Once you’ve done this, you need to verify that the $\phi$ you defined is one-to-one, onto, and a homomorphism. You will need to use 4.28; do you see why?
- As you start working with \(\mathbb{Z}_n\) and \(U_n\), remember that the operation is addition for \(\mathbb{Z}_n\) but multiplication for \(U_n\). Also, one small trick for working modulo $n$ is that you can always add or subtract $n$ without changing the object because \(n\) behaves like 0 when working mod \(n\). For example, \(5,7\in U_{12}\), so \(5\cdot 7 \equiv 35 \equiv 35 - 12 - 12 \equiv 11\) (mod $12$). Thus, in $U_{12}$, $5\cdot 7 = 11$. Similarly, in \(\mathbb{Z}_{12}\), \(5+7 \equiv 12 \equiv 12-12 \equiv 0\) (mod $12$), so $5+7 = 0$ in \(\mathbb{Z}_{12}\). Feel free to look at other resources (e.g. https://en.wikipedia.org/wiki/Modular_arithmetic) to fill in background about modular arithmetic if needed.
Extra practice: 4.28, 4.32