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Problem 44 ($\mathbb{Z}_n$ and $U(n)$ are groups)

Show the following. You need to briefly explain why the set together with its binary operation satisfies the definition of a group.

  1. For each $n\geq 1$, the set $\mathbb{Z}_n$ is a group under addition mod $n$.
  2. For each $n\geq 2$, prove that $U(n)$ is a group under multiplication mod $n$.

Try to solve the problems above without looking up the definition of a group.

If you need to, click here to show the definition of a group.

Definition (Group)

Let $G$ be a nonempty set, and let $*$ be a binary operation on $G$, which means for every $x,y\in G$ we have $x*y\in G$ $\textbf{[Closure]}$. The structure $\mathbb{G} = (G,*)$ is called a $\textdef{group}$ if the following hold.

  1. $\textbf{[Associativity]}$ For all $x,y,z\in G$ we have $(x* y)* z = x* (y* z)$.
  2. $\textbf{[Identity]}$ There is an $e\in G$ such that for all $x\in G$ we have $x * e = e* x = x$.
  3. $\textbf{[Inverses]}$ For all $x\in G$ there is a $y\in G$ such that $x* y = y* x = e$.

We usually simply write $G$ when referring to the entire structure $\mathbb{G}=(G,*)$. The element $e$ from the second point is called the $\textdef{identity}$. The element $y$ from the third point is called the $\textdef{inverse}$ of $x$ and is usually denoted $x^{-1}$. One often simply writes $xy$ in place of $x*y$, and for every positive integer $n$, we'll write $x^n$ as shorthand for $x* x* \cdots * x$ ($n$ times).

Solution

We first show $\mathbb{Z}_n$ is a group under addition mod $n$ for all $n\geq 1$. Pick $n\geq 1$. Thus we know $\mathbb{Z}_n\{1,2,\cdots,n-1\}$. Pick $a,b\in\mathbb{Z}_n$. We have two cases, namely that $a=b$ and $a\neq b$.

For the first case, suppose $a=b$. Thus we know $a+b\mod{n}$.

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