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Problem 60 (Properties Of $\langle a \rangle$ When $a$ Has Finite Order)

Let $G$ be a group with $a\in G$. Suppose that the order of $a$ is $|a|=n$. Prove the following:

  1. We have $\langle a\rangle = \{e,a,a^2,\ldots, a^{n-1}\}$. (You are showing two sets are equal.)
  2. We have $a^i=a^j$ if and only if $i-j$ is a multiple of $n$.
  3. The order of an element equals the order of the subgroup generated by that element, namely $|a|=|\langle a\rangle|$. (How can you combine 1 and 2 to get this.)

Solution

Part 1
Let $G$ be a group with $a\in G$. We suppose that the order of $a$ is $|a|=n$. We must prove that $\langle a \rangle = \{e, a, a^2, \cdots, a^{n-1}\}$. Let $B = \{e, a, a^2, \cdots, a^{n-1}\}$. To prove $\langle a \rangle = B$ we must show $B\subseteq \langle a \rangle$ and $\langle a \rangle \subseteq B$. Recall $\langle a \rangle = \{a^n | n\in \mathbb{Z}\}$. Clearly $B\subseteq \langle a \rangle$ by definition.

We will now show $\langle a \rangle \subseteq B$. Let $b\in \langle a \rangle$. This means $b=a^k$ for some $k\in \mathbb{z}$. We will use the division algorithm to write $k=qn+r$ for some $q,r\in \mathbb{Z}$ with $0\leq r < n$. We compute $$\begin{align} b&=a^k\\ &=a^{qn+r}\\ &=a^{qn}a^r\\ &=(a^n)^q a^r\\ &=e^q a^r\\ &=e a^r\\ &=a^r.\end{align}$$

Recall $0\leq r < n$. Hence we know $b\in B$ and have shown $\langle a \rangle = B$.

Part 2
We now will prove that we have $a^i=a^j$ if and only if $i-j$ is a multiple of $n$. We first suppose $a^i=a^j$. Using the division algorithm we can write $i-j=qn+r$ for some $q,r\in \mathbb{Z}$ with $0\leq r < n$. We must prove that $r=0$. We compute $$\begin{align} e &=a^i(a^i)^{-1}\\ &=a^i(a^j)^{-1}\\ &=a^i a^{-j}\\ &=a^{i-j}\\ &=a^{qn+r}\\ &=a^{qn}a^r\\ &=(a^n)^q a^r\\ &=e^q a^r\\ &=e a^r\\ &=a^r.\end{align}$$

Recall $r\in \mathbb{Z}$ with $0\leq r < n$. This means $r\in \{e, a, a^2, \cdots, a^{n-1}\}$. Since $n$ is the order of $a$, we know $n$ is the smallest positive integer such that $a^n=e$. The only option for $r$ is $0$ because if $r$ is positive the $a^r\neq e$. This proves the only if portion.

To prove the if portion, we suppose that $i-j$ is a multiple of $n$. This means $i-j=qn$ for some $q\in \mathbb{Z}$. We compute $$\begin{align} a^i &=a^i e\\ &=a^i (a^{-j} a^j)\\ &=(a^i a^{-j}) a^j\\ &=a^{i-j} a^j\\ &=a^{qn} a^j\\ &=(a^n)^q a^j\\ &=e^q a^j\\ &=e a^j\\ &=a^j. \end{align}$$ Hence $a^i=a^j$.

Part 3
Finally, we will prove that the order of an element equals the order of the subgroup generated by that element, namely $|a|=|\langle a \rangle|$. We will pick $x,y\in B$ such that $x=a^i$ and $y=a^j$ for some $i,j \in \{0, 1, 2, \cdots, n-1\}$ and $i>j$. From part 2 we know $a^i=a^j$ if and only if $i-j$ is a multiple of $n$. We will suppose $i\neq j$ and prove that $x\neq y$. Since $i\neq j$, then $i-j\neq 0$. We use the division algorithm to write $i-j=qn+r$ for some $q, r \in \mathbb{Z}$ with $0\leq r < n$. We know $i-j\in \{1, 2, \cdots, n-1\}$ since $i\neq j$. Since there are no multiples of n in $\{1, 2, \cdots, n-1\}$ we know $i-j$ is not a multiple of $n$. By part 2's contrapositive we know $a^i\neq a^j$. Hence $x\neq y$. This means there are no repeated elements in $\langle a \rangle=\{e, a, a^2, \cdots, a^{n-1}\}$ and thus there are $n$ elements. $\square$

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