> [!NOTE] Theorem (Unique inverses) > Let $(G,\cdot)$ be a [[Groups|group]] then for all $a\in G,$ $a$ has a unique [[Inverse of Group Element|inverse]]. ^5bd480 *Proof*. Suppose $e\in G$ is the identity element and $b,c\in G$ are both inverses for $a.$ Then $a\cdot b=b\cdot a=e \tag{1}$and $a\cdot c = c \cdot a = e \tag{2}$Then $\begin{align} b &= b \cdot e &\text{definition of the identity } \\ &= b \cdot ( a \cdot c) &\text{substituting for $e$ using $(2)$} \\ &= (b\cdot a)\cdot c &\text{associativity} \\ &= e \cdot c &\text{substituting for $b\cdot a$ using $(1)$} \\ & = c & \text{ definition of the identity} \end{align}$