\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
abstract banner
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \newcommand{\qed} { {\scriptstyle \Box} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::Algebra::Aluffi

Groups. 3 basic lemmas.

This card covers three very basic and fundamental properties of groups.

The identity element is unique

If \( h \in G \) is an identity of \( G \), then \( h = e_G \).

Proof. Let \( h \) and \( e_G \) be identities of \( G \). Then we have:

\[ h = e_G h = e_G.\]

The inverse is unique

If \( h_1, h_2 \) are both inverses of \( g \) in \( G \), then \( h_1 = h_2\).

Proof. Let \( h_1, h_2 \) be inverses of \( g \). Then we have:

\[ \begin{align*} (h_1 g) h_2 &= h_2 \\ h_1 (g h_2) &= h_1 \end{align*} \]

By associativity, \( (h_1 g) h_2 = h_1 ( g h_2) \), so we must have \( h_1 = h_2 \).

Cancellation

Let \( G \) be a group, and let \( a, g, h \in G \). The following holds:

\[ ga = ha \implies g = h, \quad ag = ah \implies g = h\]

Both cancellation statements follow easily by composing \( a^{-1} \) and applying associativity. To appeal to intuition, note that (I think!) an isomorphism must be both monomorphic and epimorphic (be careful to note that the inverse implication doesn't hold). Being monomorphic, \( a \) doesn't allow any morphism to "hide" after \( a \), like \( ga, ha \). Being epimorphic, \( a \) doesn't allow any morphism to "hide" before \( a \), like \( ag, ah \).


Groups as pointed sets

Aluffi mentions that the first property implies that groups can be considered to be pointed sets. A function \( \{ * \} \to G \) from a singleton to \( G \) that selects the identity element has enough information to store both the set \( G \) itself and the information about which element is the identity. My thought is though, surely the set \( G \) is sufficient alone as a datum, as the identity will be present and doesn't need to be "pointed out".

Fail to cancel

There are many set-operation pairs that do not satisfy cancellation, and thus cannot form groups. I think this statement is equivalent to saying, not every element has an inverse. \( (\mathbb{R}, \times ) \), multiplication on the reals doesn't form a group, as 0 doesn't have an inverse (can't be canceled). \( (\mathbb{R} \setminus \{ 0 \}, \times) \) does form a group though.

Source

p43