Basic Theorems Regarding the Order of Elements in a Group

Recall from 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 $a^m = e$ denoted $\mathrm{ord}(a) = m$, and if not such positive integer exists then $\mathrm{ord}(a) = \infty$.

We will now look at some basic results regarding the order of elements in a group.

Proposition 1: Let $(G, \cdot)$ be a group. Then: a) $\mathrm{order}(a) = 1$ if and only if $a = e$ where $e$ is the identity element of $G$. b) If $a \in G$ then $\mathrm{ord}(a^{-1}) = \mathrm{ord}(a)$. c) If $a, b \in G$ then $\mathrm{ord}(a \cdot b) = \mathrm{ord}(b \cdot a)$. _

• Proof of a) $\Rightarrow$ Suppose that $\mathrm{ord}(a)$. Then $a^1 = e$. So $a = e$.

• $\Leftarrow$ The smallest positive integer such that $e^n = e$ is $n = 1$. So $\mathrm{ord}(e) = 1$. $\blacksquare$

• Proof of b) Let $a \in G$. Suppose that $a$ has finite order, say $\mathrm{ord}(a) = n$. Then $n$ is the smallest positive integer such that $a^n = e$. So $a^{-n} = e$. So $\mathrm{ord}(a^{-1}) \leq n$. If $\mathrm{ord}(a^{-1}) = m < n$ then $a^{-m} = e$ implies that $a^m = e$, and since $m < n = \mathrm{ord}(a)$ we have arrived at a contradiction. Thus $\mathrm{ord}(a^{-1}) = \mathrm{ord}(a)$.

• Now suppose that $a$ has infinite order. If $a^{-1}$ has finite order, say $\mathrm{ord}(a^{-1}) = m < \infty$ then $a^{-m} = e$ implies that $a^m = e$ - contradicting $a$ having infinite order. Thus $a^{-1}$ must also have infinite order.

• Proof of c) Let $a, b \in G$. First, suppose that $\mathrm{ord}(a \cdot b) = n < \infty$. Then $n$ is the smallest positive integer such that $(a \cdot b)^n = e$. So $(a \cdot b)^n \cdot a = a$. Now observe that:

(1)

\begin{align} \quad (a \cdot b)^n \cdot a = \underbrace{(a \cdot b) \cdot (a \cdot b) \cdot ... \cdot (a \cdot b)}_{n \: \mathrm{factors}} \cdot a = a \cdot \underbrace{(b \cdot a) \cdot (b \cdot a) \cdot ... \cdot (b \cdot a)}_{n \: \mathrm{factors}} = a \cdot (b \cdot a)^n \end{align}

• Therefore $(a \cdot b)^n \cdot a = a \cdot (b \cdot a)^n$. But $(a \cdot b)^n = e$, so $a = a \cdot (b \cdot a)^n$. Thus $(b \cdot a)^n = e$. So $\mathrm{ord}(b \cdot a) \leq n = \mathrm{ord}(a \cdot b)$.

• By symmetry, we see that $\mathrm{ord}(a \cdot b) \leq \mathrm{ord}(b \cdot a)$. Thus $\mathrm{ord}(a \cdot b) = \mathrm{ord}(b \cdot a)$.

• Now suppose that $\mathrm{ord}(a \cdot b)$ is infinite. If $\mathrm{ord}(b \cdot a) = n < \infty$ then $(b \cdot a)^n = e$. By the same argument above, we see that $(a \cdot b)^n = e$ - contradicting $a \cdot b$ having infinite order. Thus $\mathrm{ord}(b \cdot a)$ is infinite. $\blacksquare$