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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

  1. 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

  2. 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?

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    *All* of $M_2(\mathbb{R})$ is a (multiplicative) monoid. If you remove the zero matrix, you still have a monoid, but it is still not a group since it has noninvertible elements. Affixing $^\ast$ refers to the invertible elements only, which does form a group.2012-10-10

2 Answers 2

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A group $G$ is defined as abelian if $AB=BA$ for all $A,B \in G$. In the case of $G=M^*_2(\mathbb{R})$, this is not true (as the example you cite shows).

It's not hard to find examples of elements $A,B \in G$ that commute in any arbitrary group $G$ (whether or not it's abelian), such as when $A=B$, when $A$ is the identity or when $A=B^{-1}$. But, when determining whether or not $G$ is abelian, the important point is whether or not this is true for all pairs of elements $A,B \in G$.

As for the second part, I'm guessing your professor wrote $M_2(\mathbb{R})$ (i.e., $2 \times 2$ matrices over $\mathbb{R}$) is a monoid (since it contains non-invertible elements -- but satifies the other group axioms under matrix multiplication). You don't want to throw away the $0$-matrix, since then it wouldn't be closed under multiplication, e.g. \[\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}\right)\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array}\right)=\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}\right).\]

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    $M_2^*(\mathbb{R})$ should be where we throw out the non-invertible elements, as per the definition in your question (in $M_2^*(\mathbb{R})$, we throw out the all $0$-matrix, along with every other non-invertible matrix).2012-10-10
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Abealian means that $AB=BA$ for all $A.B$. Thus nonabelian means $AB\ne BA$ for some $A.B$. Especially, you may (and actually do) have $AI=IA$ and $A^{-1}A=AA^{-1}$ in a nonabelian group.

The notation in question is indeed sometimes used confusingly. I had preferred $M^\times$ instead of $M^*$ for what your book writes, where $M^\times $ explicitly denotes the units of a ring, i.e. the elements that happen to have a multiplicative inverse. There are simply too many things that can be denoted by adding a star.

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    I haven't learned what a ring is. I am sorry2012-10-10