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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.

p43