inverse of a productInverseFormingInProportionToGroupOperation2004-12-13 15:42:162008-05-10 12:34:55Theoremgroupinversegroup operation% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}\begin{thmplain}
\,If $a$ and $b$ are arbitrary elements of the group \,$(G,\,*)$, then the inverse of $a*b$ is
\begin{align}
(a*b)^{-1} = b^{-1}*a^{-1}.
\end{align}
\end{thmplain}
{\em Proof.}\, Let the neutral element of the group, which may be proved unique, be\, $e$.\, Using only the group postulates we obtain
$$(a*b)*(b^{-1}*a^{-1}) = a*(b*(b^{-1}*a^{-1})) =
a*((b*b^{-1})*a^{-1}) = a*(e*a^{-1}) = a*a^{-1} = e,$$
$$(b^{-1}*a^{-1})*(a*b) = b^{-1}*(a^{-1}*(a*b)) = b^{-1}*((a^{-1}*a)*b) =
b^{-1}*(e*b) = b^{-1}*b = e,$$
Q.E.D.
\textbf{Note.}\, The \PMlinkescapetext{formula} (1) may be by induction extended to the form
$$(a_1*\cdots*a_n)^{-1} = a_n^{-1}*\cdots*a_1^{-1}.$$