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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

1. Clearly $\{e,a,a^2,\cdots,a^{n-1}\} \subseteq \langle a\rangle$. We now show $\langle a\rangle \subseteq \{e,a,a^2,\cdots,a^{n-1}\}$. Let $x\in \langle a \rangle$. This means there exists $y\in \mathbb{Z}$ such that $x=a^y$. By the division algorithm, there exist unique $q,r \in \mathbb{Z}$ such that $y=qn+r$, where $0\leq r<n$. We compute

$$ x=a^y=a^{qn+r}=a^{qn}a^{r}=(a^n)^qa^r=(e)^qa^r=a^r $$

Thus $x=a^r$, where $0\leq r\leq n-1$, and hence $x \in \{e,a,a^2,\cdots,a^{n-1}\}$.

2. Suppose $a^i=a^j$. We will show $i-j$ is a multiple of $n$. By the division algorithm, there exists unique $q_1,r_1 \in \mathbb{Z}$ such that $i=q_1n+r_1$, where $0\leq r_1<n$, and there exists unique $q_2,r_2 \in \mathbb{Z}$ such that $j=q_2n+r_2$, where $0\leq r_2<n$. Since $a^i=a^j$, we know $a^{q_1n+r_1}=a^{q_2n+r_2}$, which simplifies to $(a^{n})^{q_1}a^{r_1}=(a^{n})^{q_2}a^{r_2}$. Since $a$ has order $n$, it follows that $a^{r_1}=a^{r_2}$. Multiplying by $a^{-r_2}$ on the right, we obtain $a^{r_1-r_2}=e$. Since $0 \leq r_1,r_2 \leq n-1 $, it follows that $-(n-1)\leq r_1-r_2\leq n-1$. Using this fact, along with the fact that $n$ is the smallest positive integer $k$ such that $a^k=e$, it follows that $r_1-r_2=0$. It then follows that $i-j= q_1n+r_1 - (q_2n+r_2) = (q_1-q_2)n+(r_1-r_2)=(q_1-q_2)n$, or $i-j$ is a multiple of $n$.

Now suppose $i-j$ is a multiple of $n$. We must show $a^i=a^j$. We know $i-j=qn$ for some $q \in \mathbb{Z}$. Thus $a^{i-j}=a^{qn}=(a^{n})^q=(e)^q=e$. Multiplying both sides on the right by $a^j$, we obtain $a^{i-j}a^j=ea^{j}$, which simplifies to $a^i=a^j$.

3. We must show $|\langle a\rangle |=n$. We know from part 1 that $|\langle a\rangle |\leq n$. Let $x,y \in \langle a\rangle$. Pick $i,j$ such that $x=a^i$ and $y=a^j$. Since $0\leq i,j \leq n-1$, we know $-(n-1)\leq i-j \leq n-1$. Hence if $i-j$ is a multiple of $n$, then $i-j=0$, or $i=j$. Hence we know that $a^i =a^j$ if and only if $i=j$. This implies $|\langle a\rangle |\geq n$. We conclude that $|\langle a\rangle |=n$.

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