Distinct Powers of Elements in a Group

Recall from the The Order of Elements in a Group page that if $(G, \cdot)$ is a group and $a \in G$ then the order of $a$ is the least positive integer $m$ such that $a^m = (\underbrace{a \cdot a \cdot … \cdot a}_{m \: \mathrm{many \: factors}}) = e$ where $e$ is the identity element, and if no such positive integer $m$ exists then $a$ is said to have order $/infty$.

We will now look at a rather nice proposition which will tell us that if $a$ has order $m$, then the nonnegative powers of $a$ less than $m$ are all distinct elements in the group.

Proposition 1: Let $(G, \cdot)$ be a group and let $a \in G$ have order $m$. Then the nonnegative integer powers less than $m$ are distinct, that is $a^0, a^1, …, a^{m-1}$ and these are the only powers of $a$.

• Proof: Let $(G, \cdot)$ be a group and let $a \in G$ have order $m$. Then $m$ is the least positive integer such that $a^m = e$ where $e$ is the identity element of $G$. We will begin by showing that these powers $a^0, a^1, …, a^{m-1}$ are the other powers of $a$. Let $s$ be any nonnegative integer. By The Division Algorithm we have that for integers $q$ and $r$ such that $0 \leq r < m$ that represent the quotient and remainder of $s$ upon division by $m$ that:

(1)

\begin{align} \quad s = mq + r \end{align}

• Therefore:

(2)

\begin{align} \quad a^s = a^{mq + r} = a^{mq} \cdot a^r = (a^m)^q \cdot a^r = e^q \cdot a^r = e \cdot a^r = a^r \end{align}

• Hence $a^s = a^r$, but $r \in \{ 0, 1, …, m - 1 \}$ which proves that $a^0, a^1, …, a^{m-1}$ are the only powers of $a$.

• We will now show that the powers $a^0, a^1, …, a^{m-1}$ are distinct. Suppose that for some $s, t \in \{ 0, 1, …, m-1 \}$ we have that:

(3)

\begin{align} \quad a^s = a^t \end{align}

• Without loss of generality, assume that $s \geq t$. Then:

(4)

\begin{align} \quad a^{s - t} = e \end{align}

• Since $s, t < m$ we must have that $0 \leq s - t < m$. If $s - t \neq 0$ then this implies that the order of $a$ is $s - t < m$ which is a contradiction. Therefore $s - t = 0$ or rather, $s = t$. Hence $a^s = a^t$ if $s = t$, so the powers $a^0, a^1, …, a^{m-1}$ are distinct. $\blacksquare$

Proposition 2: Let $(G, \cdot)$ be a group and suppose that $a \in G$ has order $\infty$. Then all powers of $a$ are distinct, that is $a^0, a^1, a^2, …$ are all distinct.

• Proof: Let $(G, \cdot)$ be a group and suppose that $a \in G$ has order $\infty$. Let $s$ and $t$ be nonnegative integers and suppose that:

(5)

\begin{align} \quad a^s = s^t \end{align}

• Without loss of generality, assume that $s \geq t$. Then:

(6)

\begin{align} \quad a^{s - t} = e \end{align}

• If $s - t > 0$ then this implies that $a$ has order $s - t$ which contradicts the hypothesis that $a$ has order $/infty$. Hence $s - t = 0$, so $s = t$. Therefore all of the powers $a^0, a^1, a^2, …$ of $a$ are unique. $\blacksquare$

Corollary 3: If $(G, \cdot)$ is a finite group then for all $a \in G$, $a$ has a finite order.

• Proof: Let $(G, \cdot)$ be a group and suppose that for some $a \in G$ that $a$ has order $\infty$. Then all of the powers $a^0, a^1, a^2, …$ are distinct by Proposition 2. Since $(G, \cdot)$ is closed under $\cdot$ we must have that all of these powers are contained in $G$ contradicting the hypothesis that $(G, \cdot)$ is a finite group. Hence every $a \in G$ has a finite order. $\blacksquare$