📂Abstract Algebra

Proof of the Law of Cosines

Theorem ¹

For an element $a,b,c$ of a group $\left<G, \ast \right>$, $$ a \ast b = a \ast c \implies b = c \\ b \ast a = c \ast a \implies b=c $$

Explanation

When one encounters abstract algebra, they begin to learn what they have previously known in a new language. The cancellation property is probably among the first theorems one encounters. Normally, we simply say that we divide (multiply by the inverse) the same thing on both sides. The term “cancellation property” is used in Japan, and if one wishes to remember it by any name, it would be good to be aware of the expression used in English.

Proof

If we say $a \ast b = a \ast c$, since $a$ is an element of $G$, there exists a left inverse $a '$. Multiplying both sides by the left inverse $a '$ gives $$ a’ \ast\ (a \ast b) = a’ \ast\ (a \ast c) $$ Since the associative law holds, $$ (a’ \ast\ a) \ast b = (a’ \ast\ a) \ast c $$ By the definition of the inverse, if the identity of $G$ is called $e$, then it is $(a’ \ast\ a) = e$, therefore $$ e \ast b = e \ast c $$ By the definition of identity, $$ b = c $$ The proof for the right-hand side can be done in the same manner.

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