Show the set of units $R^\times$ in $R$ forms a group 
Let $R$ be a ring with unity. Show that the set of units $R^\times = \{r ∈ R\mid \exists r^{−1} \in R, rr^{−1} = r^{−1}r = 1\}$ in $R$ forms a group.


proof:
Since $R$ is a ring, $\exists 1\in R$ such that $1\cdot 1^{-1}=1^{-1}\cdot 1=1$ , hence $R^X\neq \{\phi\}$.
let $a,b,c\in R^X$, then $a=r_1r_1^{-1}, b=r_2r_2^{-1},c=r_3r_3^{-1}$ in $R$ where $r_i\in R$ and $i=1,2,3$, thus $(a\cdot b)\cdot c=r_1r_1^{-1}\cdot r_2r_2^{-1}\cdot r_3r_3^{-1}=1=a\cdot(b\cdot c)$
And $\exists a^{-1}=(r_1r_1^{-1})^{-1}\in R$ since $R$ is a ring, then
$\begin{align} aa^{-1}&= (r_1r_1^{-1})(r_1r_1^{-1})^{-1} \\ &=(r_1r_1^{-1}) (r_1^{-1}r_1)\\&=(r_1r_1^{-1}) (r_1r_1^{-1})\\&=r_1(r_1^{-1} r_1)r_1^{-1}\\ &=1\end{align}$
Hence $R^X$ is a group under multiplication. 

Can anyone check where I did incorrect?  Thanks
 A: It is clear that $R^{\times}$ is a non-empty set, since $1 \in R$ ($R$ is a ring with unity), and:
$1\cdot 1 = 1$, that is, $1 \in R^{\times}$.
So, $R^{\times}$ is a group if and only if it is closed (under the multiplication of $R$), and if it possesses all inverses.
Closure is easy to show: let $a,b \in R^{\times}$. This means there exists $a',b' \in R$ such that:
$aa' = a'a = 1$, and $bb' = b'b = 1$. So now we need to find $c \in R$ such that:
$(ab)c = c(ab) = 1$. Clearly, $c = b'a'$ will do the trick:
$(ab)(b'a') = a(bb')a' = a(1)a' = aa' = 1$, while:
$(b'a')(ab) = b'(a'a)b = b'(1)b = b'b = 1$ (note how we used associativity of the multiplication of $R$, and the fact that $1$ is a multiplicative identity).
We conclude that $ab \in R^{\times}$ whenever $a,b$ are.
The existence of inverses is also fairly trivial. Suppose $a \in R^{\times}$. Then there exists $a' \in R$ with $aa' = a'a = 1$. So it seems natural to let $a^{-1} = a'$. However, we need to show $a'$ is indeed already in $R^{\times}$, not just in $R$.
But clearly, there is $(a')' \in R$ such that $a'(a')' = (a')'a' = 1$, namely $(a')' = a$ will do. So $R^{\times}$ indeed possesses all inverses, and thus forms a group (and moreover, we can now conclude the inverse of a unit is unique).

Your argument that $a = r_1r_1^{-1}$ makes no sense at all: such an $a$ is NOT arbitrary, since $r_1r_1^{-1} = 1$ (should the element $r_1^{-1}$ even exist at all).
