The identity element and the inverse A group is a ordered pair $(G,*)$ which satisfies the following,


*

*$G$ is a set

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

*$*$ is associative

*$\exists e\in\mathbb{G} (\forall a\in\mathbb{G} a*e=e*a=a)$ 

*$\forall a\in\mathbb{G}\exists b\in\mathbb{G}(a*b=b*a=e)$
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
 A: 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 $a\in G$ 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 $a^{-1}(aa') = ???$.
A: 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.
A: 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?
A: 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.
A: 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:
$$
y_1=y_1*e=y_1*(x*y_2)=(y_1* x) * y_2=e* y_2=y_2
$$
