Bdtyktl

A group is a ordered pair $(G,*)$ which satisfies the following,

1. $G$ is a set




2. $*$ is a binary operation on $G$




3. $*$ is associative




4. $exists einmathbb{G} (forall ainmathbb{G} a*e=e*a=a)$




5. $forall ainmathbb{G}exists binmathbb{G}(a*b=b*a=e)$

Note : $e$ is called THE identity element of this group $(G,*)$.

Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.

Prove "THE", in other words, show identity is unique and inverse is unique.

I need a little bit help maybe and idea on how to start such problem.

Thank you

Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.

Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?

As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.

What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.

Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.

Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?

If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?

Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.

Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.

Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.

If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2