\(
\newcommand{\cat}[1] {\mathrm{#1}}
\newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})}
\newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}}
\newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}}
\newcommand{\betaReduction}[0] {\rightarrow_{\beta}}
\newcommand{\betaEq}[0] {=_{\beta}}
\newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}}
\newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}}
\newcommand{\groupMul}[1] { \cdot_{\small{#1}}}
\newcommand{\groupAdd}[1] { +_{\small{#1}}}
\newcommand{\inv}[1] {#1^{-1} }
\newcommand{\bm}[1] { \boldsymbol{#1} }
\require{physics}
\require{ams}
\require{mathtools}
\)
Math and science::Algebra::Aluffi
Groups. All orders divides order of maximal element
Order: divides order of maximal element
Let \( G \) be a [something] group, and let \( g \in G \)
be an element with maximal finite order. Any other element with
finite order has an order that [has what relationship?].
The same statement cannot be made for non-comutative groups.
Can you remember the proof?
Maximal finite order
Let \( g \) be an element of group \( G \). \( g \) has
maximal finite order iff all other elements with finite order
have an order less than that of \( g \).
