Basic Theorems Regarding the Order of Elements in a Group

# Basic Theorems Regarding the Order of Elements in a Group

Recall from The Order of Elements in a Group page that if $(G, *)$ 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 theorems regarind the order of elements in a group.

 Theorem 1: Let $(G, *)$ be a group and let $a \in G$. Then $\mathrm{order}(a) = 1$ if and only if $a = e$ where $e$ is the identity element with respect to $*$.
• Proof: $\Rightarrow$ Suppose that $a \in G$ is such that $\mathrm{ord}(a) = 1$. Then $a^1 = a = e$ so $a = e$.
• $\Leftarrow$ Now suppose that $a = e$. Then $a^1 = e$ so $\mathrm{ord}(a) = 1$. $\blacksquare$
 Theorem 2: Let $(G, *)$ be a group and let $a \in G$. Then if $\mathrm{ord} (a) = m$ then $\mathrm{ord} (a^{-1}) = m$.
• Proof: Let $\mathrm{ord} (a) = m$ and $\mathrm{ord}(a^{-1}) = n$ Then $a^m = e$ and $a^n = e$. Suppose that $m /neq n$. Then either $m > n$ or $m < n$. If $m > n$ then $m - n > 0$ and:
(1)
\begin{align} \quad e = a^m * (a^{-1})^n = a^{m-n} \end{align}
• Therefore $a^{m-n} = e$, but $0 < m - n < m$ which contradicts the hypothesis that $\mathrm{ord} (a) = m$. Similarly, if $m < n$ then $n - m > 0$ and:
(2)
\begin{align} \quad e = (a^{-1})^n * a^m = (a^{-1})^{n-m} \end{align}
• Therefore $(a^{-1})^{n-m} = e$, but $0 < n - m < n$ which contradicts the hypothesis that $\mathrm{ord} (a^{-1}) = n$, Therefore our assumption that $m /neq n$ was false and so $\mathrm{ord} (a) = \mathrm{ord} (a^{-1})$