Homework 16

Read: Continue with Section 4.1.

Turn in: 4.27 or 4.29 (you choose one), 4.36, 4.37, 4.38, 4.40, 4.42

• 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.

• 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 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. 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.

• On 4.40, you just need to properly link together Theorems 4.27, 4.29, and 4.39. Also, remember that Theorem 3.56 implies that if $G_1 \cong G_2$ and $G_2 \cong G_3$, then $G_1 \cong G_3$.

Extra practice: 4.28, 4.32, 4.39, 4.41

• 4.41 is quite challenging. Assuming $G$ is cyclic and $H$ is a subgroup, you aim to find a generator for $H$. Since $G$ is cyclic, it has some generator $a$. This means that the elements of $G$, hence the elements of $H$, can be written as powers of $a$, but which one might be a generator for $H$? If $H$ is the trivial subgroup, then $H$ is generated by $e$, and you are done. If $H$ is nontrivial, then one idea is to consider $a^k$ where $k$ is the smallest positive integer such that $a^k\in H$.

• Challenge problem (just for fun): Problem 4.20 implies that if $G$ is finite, then $\forall g\in G$, $\exists n\in \mathbb{N}$ such that $g^n = e$. What happens if you switch the order of the quantifiers? Try to prove this: if $G$ is finite, then $\exists n\in \mathbb{N}$ such that $\forall g\in G$, $g^n = e$. Do you see the difference?