> [!NOTE] Theorem (Distributivity over group operation) > Let $G$ be an [[Groups|Abelian group]]. Let $a,b\in G$ and $n\in\mathbb{Z}$ then using [[Multiplicative Notation|multiplicative notation]], $(ab)^{n}=a^{n}b^{n}$or using [[Additive Notation|additive notation]], $n(a+b)=na+nb$ **Proof**: For $n=0,$ the statement is is true as $(ab)^{0}=1=1\circ 1= a^{0}b^{0}.$ If $n$ is a positive integer. Then $(ab)^{n}= \underbrace{abab\cdots ab}_{n \text{ times}}=\underbrace{aaa\cdots a}_{n \text{ times}}\; \underbrace{bbb\cdots b}_{n \text{ times}}=a^{n}b^{n}$If $n$ is negative, we have $\begin{align} (ab)^{n} &= ((ab)^{-n})^{-1} & \text{definition of negative power} \\ &= ((ab)^{-1})^{-n} &\text{using $(a^{n})^{-1}= (a^{-1})^{n}$} \\ &= (b^{-1}a^{-1})^{-n} &\text{using $(ab)^{-1}=b^{-1}a^{-1}$} \\ &=(b^{-1})^{-n} (a^{-1})^{-n } &\text{using $(ab)^{n}=a^{n}b^{n}$ since $n\in\mathbb{N}$} \\ &=b^{n}a^{n} &\text{ using $(a^{-1})^{n}=a^{-n}$} \\ &= a^{n}b^{n} &\text{using commutativity of binary operation} \end{align}$ # Applications Can we construct a ring from group like this?