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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!) a 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 though 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