Hw4 Problem 7

### Problem 7

Let $G$ be a group with a finite number of elements. For $a \in G$, we define $a^n=a*a* \cdots *a$ for $n$ factors. Show that for any $a \in G$, there exists an $n\in \mathbb{Z}^+$ such that $a^n=e$. [Hint: Consider $e, a, a^2, a^3, \dots, a^m$ where $m$ is the number of elements in $G$, and use the cancellation laws.]

Solution
Let $G$ be a group with a finite number of elements, and let $a\in G$.
Now, since $G$ is a group, it is closed under the operation $*$, and since $G$ is both finite and closed under $*$, $a^k \in G$ for all $k\in \{0,1,2,...,j\}$, where $j>k$ for all $k \in \{0,1,2,...,j\}$.
Clearly, either all elements of the form $a^k$ are unique in $G$, or there exist some elements $a^x,a^y \in G$ with $x\neq y$ s.t. $a^x=a^y$.
$\hspace{20pt}$ Suppose all $a^k \in G$ are unique in $G$.
$\hspace{20pt}$ Now, consider $a^j \in G$.
$\hspace{20pt}$ Observe that $a*a^j=a^{j+1} \in G$, since $G$ is closed under $*$.
$\hspace{20pt}$ But this is a contradiction, since $j+1>j$, and it was already shown that $j>k$ for all $a^k \in G$, with $j,k \in \mathbb{Z}^+$.
$\hspace{20pt}$ Thus, there exist some elements $a^x,a^y \in G$ with $x\neq y$ s.t. $a^x=a^y$.
Now, without loss of generality, let $x>y$, and define $(a^k)'=a'*a'*...*a'$ for $k$ factors, where $a'$ is the inverse of $a$.
By definition of group, $a' \in G$, and so $(a^k)'\in G$
Then $a^x*(a^y)'=a^y*(a^y)'$.
Hence, $a^{x-y}=e$, where $x-y\in \mathbb{Z}^+$.
Therefore, all $a\in G$ have an associated $n\in \mathbb{Z}^+$ s.t. $a^n=e$.
$\hspace{200pt} \clubsuit$