| Version 2 |
Version 1 |
| \textbf{Theorem.} \,If $a$ and $b$ are arbitrary elements of the group \,$(G,\,*)$, then the inverse of $a*b$ is |
\textbf{Theorem.} \,If $a$ and $b$ are arbitrary elements of the group \,$(G,\,*)$, then the inverse of $a*b$ is |
| $$(a*b)^{-1} = b^{-1}*a^{-1}.$$ |
$$(a*b)^{-1} = b^{-1}*a^{-1}.$$ |
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| {\em Proof.} \, If the neutral element of the group is $e$, then we obtain |
{\em Proof.} \, If the neutral element of the group is $e$, then we obtain |
| $$(a*b)*(b^{-1}*a^{-1}) = a*(b*(b^{-1}*a^{-1})) = |
$$(a*b)*(b^{-1}*a^{-1}) = a*(b*(b^{-1}*a^{-1})) = |
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a*((b*b^{-1})*a^{-1}) = a*(e*a^{-1}) = a*a^{-1} = e,$$
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a*((b*b^{-1})*a^{-1}) = a*(e*a^{-1}) = a*a^{-1} = e,$$
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$$(b^{-1}*a^{-1})*(a*b) = b^{-1}*(a^{-1}*(a*b)) = b^{-1}*((a^{-1}*a)*b) =
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$$(b^{-1}*a^{-1})*(a*b) = b^{-1}*(a^{-1}*(a*b)) = b^{-1}*((a^{-1}*a)b) =
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| b^{-1}*(e*b) = b^{-1}*b = e.$$ |
b^{-1}*(e*b) = b^{-1}*b = e.$$ |
