My book wrote something like this
Consider $M_{2}^*(\mathbb{R}) = \left \{ A \in M_{2}(\mathbb{R}) : \det(A) \neq 0\right \}$
$A = \bigl(\begin{smallmatrix} 1 &1 \\ 1 &0 \end{smallmatrix}\bigr)$ and B = $\bigl(\begin{smallmatrix} 0 &1 \\ 1 &1 \end{smallmatrix}\bigr)$ and $AB \neq BA$
It follows that ($M_{2}^*(\mathbb{R})$, . ) is a nonabelian group
My confusions
If it is nonabelian, then it is not commutative. That is the identity property $AI \neq IA = A$ isn't satisified. The same goes for the inverse
In my lecture my professor just wrote ($M_{2}^*(\mathbb{R})$, . ) and told us that this is a Monoid (and not a group) because he told us that a * on top of a group means without zero (matrix here). And he wrote that because not all 2 x 2 matrices are invertible. Which confuses the notations people generally use. But what he wrote (in my notes) seem to contradict the book's example
Can someone clarify my confusion?