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Proposition:
  1. The identity is unique.
  2. Inverses are unique.
  3. \((a^{-1})^{-1} = a\)
  4. \((a \cdot b)^{-1} = b^{-1} \cdot a^{-1}\)
  5. General associativity: When computing \(a_1 \cdot a_2 \cdot \dots a_n\), the computation is invariant of the order of operation.
  6. Cancellation: If \(a \cdot u = a \cdot v\) then \(u = v\) and if \(u \cdot a = v \cdot a\) then \(u = v.\)

Proof \(\downarrow\):
  1. By way of contradiction, suppose \(e_1 \neq e_2\) and both are identities. Then \[e_1 = e_1 \cdot e_2 = e_2\] Therefore, there can only be one identity.
  2. By way of contradiction, suppose \(b \neq c\) are both inverses of \(a.\) By definition of inverse, \(a \cdot c = 1.\) By definition of identity, \begin{align} b & = b \cdot 1 \\ & = b \cdot (a \cdot c) \end{align} Now use associativity to change the order of multiplication. \begin{align} b \cdot (a \cdot c) & = (b \cdot a) \cdot c \\ & = 1 \cdot c \\ & = c \end{align} This chain of equations shows \(b = c,\) a contradiction. Therefore, inverses must be unique.
  3. Since \(a \cdot a^{-1} = 1,\) a is the unique inverse of \(a^{-1}.\) The notation of the inverse of \(a^{-1}\) is \((a^{-1})^{-1}.\)
  4. This result follows by directly checking. Begin by using associativity twice. \begin{align} (b^{-1} \cdot a^{-1}) \cdot (a \cdot b) & = b^{-1} \cdot (a^{-1} \cdot (a \cdot b)) \\ & = b^{-1} \cdot ((a^{-1} \cdot a) \cdot b) \\ & = b^{-1} \cdot (1 \cdot b) \\ & = b^{-1} \cdot b \\ & = 1 \end{align} The computation to show \((a \cdot b) \cdot (b^{-1} \cdot a^{-1}) = 1\) is similar.
  5. We will use strong induction.
    Base case: In an expression of the form \(a_1 \cdot a_2 \cdot a_3\) with just \(2\) operations, the definition of associativity tells us we can perform the operations in any order.
    Suppose we have shown that expressions with up to \(n\) operations can be performed in any order for some \(n \geq 2.\) Let \(a_1 \cdot a_2 \cdot \dots \cdot a_{n+2}\) be an expression with \(n+1\) operations. First, notice that all such ways to perform the operations with the first operation being performed last are equal by induction. \[a_1 \cdot (a_2 \dots a_{n+2})\] must yield the same result no matter what order we perform the operations in the parentheses since there are \(n\) operations within the parentheses.
    Now suppose the \(k\)th operation is performed last where \(2 \leq k \leq n+1.\) \[(a_1 \cdot \dots \cdot a_k) \cdot (a_{k+1} \cdot \dots \cdot a_{n+2})\] The number of operations on the left is \(k-1 \leq n\) so by strong induction we can perform the operations in any order and get the same result. So, we can perform the first operation last within the first set of operations: \[(a_1 \cdot (a_2 \cdot \dots \cdot a_k)) \cdot (a_{k+1} \cdot \dots \cdot a_{n+2})\] By associativity of \(\cdot,\) we can multiply the first operation last. \[a_1 \cdot ((a_2 \cdot \dots \cdot a_k) \cdot (a_{k+1} \cdot \dots \cdot a_{n+2}))\] Now the first operation is performed last, so this expression yields the same result of all orders of operations in which the first operation is performed last. Therefore, all orders in which the operations are performed yield the same results.
  6. Suppose \(a \cdot u = a \cdot v.\) Multiplying both sides on the left by \(a^{-1}\) gives \[a^{-1} \cdot (a \cdot u) = a^{-1} \cdot (a \cdot v)\] By associativity, this becomes \begin{align} & (a^{-1} \cdot a) \cdot u = (a^{-1} \cdot a) \cdot v \\ & 1 \cdot u = 1 \cdot v \\ & u = v \end{align} The argument for cancelling \(a\) on the right in \(u \cdot a = v \cdot a\) is similar.