Prove that any non empty set G with a binary operation $\bullet$ satisfying the following requirements is a group Prove that any non empty set G with a binary operation $\bullet$ satisfying the following requirements is a group:


*

*If $x,y \in G$, then $x \bullet y \in G$.

*$\forall x,y,z \in G, (x \bullet y) \bullet z=x \bullet (y \bullet z)$

*$\forall a,b \in G$, the equations $x \bullet a=b$ and $a \bullet
    y=b$ both have solutions


Hint: 


*

*Show that if $a \in G$, then a has a right and left identity and be sure to show they are the same.

*Show that the identity in step 1 works for any $b \in G$


$\textbf{Note:}$ Steps 1 and 2 can be reversed


*Show that for any $a \in G$ a has a left and right inverse and they are the same.


$\textbf{Step 1:}$ Show that if $a \in G$, then a has a right and left identity and be sure to show they are the same.
$\textbf{Definitions:}$


*

*An element $e \in G$ is called a $\textbf{left identity}$ element if $e
   \bullet x=x$ for all $x \in G$.

*An element $e \in G$ is called a $\textbf{right identity}$ element if
$x \bullet e=x$ for all $x \in G$.


$\textbf{Proof:}$ Assume that $a \in G$. Then we need to show that  (1) $e \bullet a=a$ (left identity), (2) $a \bullet e=a$ (right idenity), and (2) $e \bullet a=a=a \bullet e$
Am I reading the problem correctly? Also it seems to me odd to prove this because of the definition of identity.
 A: $\textbf{Definitions:}$


*

*An element $e \in G$ is called a \textbf{left identity} element if $e
   \bullet x=x$ for all $x \in G$.

*An element $e \in G$ is called a \textbf{right identity} element if
$x \bullet e=x$ for all $x \in G$.


$\textbf{Proof:}$ We need to show that the set G with the binary operation, $\bullet$, is a group. For G to be a group with the operation $\bullet$, we must show that it has the following items: (1) Identity, (2) Inverses, (3) Associative, and (4) Closure.
From the requirements given, there is no need to prove that G is associative nor G has closure because they are given. (i.e. Requirement A implies Closure, and Requirement B implies Associativity.) So our proof will only consist of us showing that the set G with the operation $\bullet$ has (1) a unique identity element and (2) that for each element there is only one unique inverse.
$\textbf{Proof of $(G, \bullet)$ has a unique identity:}$ To prove that $(G, \bullet)$ has a unique identity we are going to use the Hint above. We don't know that e, the identity element, exists in G. By requirement 3, we know that $ax=a$ and $ya=a$ for some $x,y \in G$. So yes there is a left and right inverse, but now we need to show that x=y. Hence we must take the following cases:


*

*If $ax=y$ and $1=a$, then $y=ax=1x=x$.

*If $ax=a$ and $y=1$, then $x=1=y$.

*If $a=a$ and $x=y$, then $x=y$.

*If $a=y$ and $x=a$, then $x=y$.


Hence we have showed that a has an identity element. We now need to show that all elements in G will have an identity element. 
Let $ b\in G$ be an arbitrary element. If we follow the proof above we will get that b will have an identity element. Since b is an arbitrary element in G, then all the elements in G have an identity element.
Note that the identity element must be unique. The proof of this claim is quite trival. Assume $e,e' \in G$ are distinct identity elements.
$$eg=ge=g \,and \,e'g=ge'=g$$
But \begin{equation*}
\begin{aligned}
ge=g=ge' & \iff e=e' \\
eg=g=e'g & \iff e=e' \\
\end{aligned}
\end{equation*}
So $e=e'$ is unique.
$\textbf{Proof that for each element in $(G, \bullet)$ there is only one unique inverse:}$ We know from the previous proof that G has an identity element, let's call e. We need to show that every element of $(G, \bullet)$ has a unique inverse. Assume $g_1,g_2 \in G$ are distinct inverse elements for an arbitrary $g \in G$.
Assume $gg_1=g_1g=e$ and $gg_2=g_2g=e$. 


*

*If $gg_1=e$, then $(gg_1)g_2=eg_2=g_2$.

*If $gg_2=e$, then $g_1(gg_2)=g_1e=g_1$.


So $g_1=g_1(gg_2)=(g_1g)g_2=g_2$ is unique.
Hence $(G, \bullet)$ is a group.
