The Laws of Exponents

# The Laws of Exponents

Just like in regular arithmetic, the laws of exponents also holds on groups $(G, *)$ under their defined operation $*$ as we will state and prove in the following theorems.

Theorem 1: Let $(G, *)$ be a group and let $a \in G$. If $m$ is a positive integer then $(a^m)^{-1} = (a^{-1})^m$ |

**Proof:**Let $a \in G$ and let $m$ be a positive integer. Then:

\begin{align} \quad (a^m)^{-1} = (\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}})^{-1} = \underbrace{a^{-1} * a^{-1} * ... * a^{-1}}_{m \: \mathrm{many \: factors}} = (a^{-1})^m \quad \blacksquare \end{align}

Theorem 2: Let $(G, *)$ be a group and let $a \in G$. If $m$ and $n$ are integers then $a^m * a^n = a^{m+n}$. |

**Proof:**Let $a \in G$ and let $m$ and $n$ be integers. Then:

\begin{align} \quad a^m * a^n = (\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}}) * (\underbrace{a * a * ... * a}_{n \: \mathrm{many \: factors}}) = \underbrace{a * a * ... * a}_{m+n \: \mathrm{many \: factors}} = a^{m+n} \quad \blacksquare \end{align}

As a consequence of Theorems 1 and 2 we will often write $a^{-m} = (a^m)^{-1} = (a^{-1})^m$ and $a^0 = a^{1} * a^{-1} = e$ where $e$ is the identity element of the specified group.

Theorem 3: Let $(G, *)$ be a group and let $a \in G$. If $m$ and $n$ are integers then $(a^m)^n = a^{mn}$. |

**Proof:**Let $a \in G$ and let $m$ and $n$ be integers. Then:

\begin{align} \quad (a^m)^n = (\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}})^n = \underbrace{(\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}})*(\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}})*...*(\underbrace{a * a * ... * a}_{m \: \mathrm{many \: factors}})}_{n \: \mathrm{many \: factors}} = a^{mn} \quad \blacksquare \end{align}