Lecture 02

Lemma

Let $G$ be a group. Then:

1. The identity of $G$ is unique;
2. if $a\in G$ then $a^{-1}$ is unique;
3. $(a^{-1}{)}^{-1}=a$
4. $(ab{)}^{-1}=b^{-1}{a}^{-1}$

Proof

$\boxed{1}$ Suppose $\exists e,f\in G$ s.t they are both identities

$e\underline{f}=e$

and

$\underline{e}f=f$

$⟹e=f$

$\boxed{2}$ Suppose $b,c$ are both inverses of $a$.

$(ba)c=ec=c$

$b(ac)=be=b$

$⟹b=c$

$\boxed{3}a{a}^{-1}=a^{-1}a=e⟹(a^{-1}{)}^{-1}=a$

$\boxed{4}$ WTS $(ab{)}^{-1}=b^{-1}{a}^{-1}$

$(ab)(b^{-1}{a}^{-1})=a(b{b}^{-1})a^{-1}=ae{a}^{-1}=a{a}^{-1}=e$

$(b^{-1}{a}^{-1})(ab)=b^{-1}(a^{-1}a)b=e$

Theorem: Cancellation Law

Let $G$ be a group and $a,b,c\in G$.

If either:

$ab=ac$ or$ba=ca$

then

$b=c$

Proof

if $ab=ac$

$(a^{-1}a)b=(a^{-1}a)c$

$eb=ec$

$b=c$

Corollary

Let $G$ be a group, $a,b\in G$. Then:

There exists exactly one $c\in G$ such that $ac=b$

and exactly one $d\in G$ such that $da=b$.

(Examples drawing out Cayley Tables in class I don’t want to type ts)