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{aligned} (h_{1} g) h_{2} & = h_{2} \\ h_{1} (g h_{2}) & = h_{1} \end{aligned}

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:

g a = h a ⟹ g = h, a g = a h ⟹ 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 $g a, h a$ . Being epimorphic, $a$ doesn't allow any morphism to "hide" before $a$ , like $a g, a h$ .

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. $(R, \times)$ , multiplication on the reals doesn't form a group, as 0 doesn't have an inverse (can't be canceled). $(R ∖ {0}, \times)$ does form a group though.

Source

p43