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'inverse of a product'
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| Title of object: |
inverse of a product |
| Canonical Name: |
InverseFormingInProportionToGroupOperation |
| Type: |
Theorem |
| Created on: |
2004-12-13 15:42:16 |
| Modified on: |
2008-04-23 17:59:13 |
| Classification: |
msc:20-00, msc:20A05 |
| Keywords: |
inverse, group operation |
| Synonyms: |
inverse of a product=inverse of a product in group |
Revision comment (for changes between this and next version):
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem} |
Content:
\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
$$(a_1*\cdots*a_n)^{-1} = a_n^{-1}*\cdots*a_1^{-1}.$$ |
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