Monoid but not a group

Question

so I am studying in Germany and I stumbled upon this kind of question so I have this set given

$$M :=\begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix}$$
$a$ and $b$ are elements of the natural numbers

Now I have to show that $(M, ·)$ is a monoid but not a group. How should I do that?

Answer 1

For $a=1$ and $b=1$ we get an identity element $e$.

The associative property is obvious because we have it in $\mathbb N$.

Thus, it's indeed a monoid.

But for $x\in M$ there is a problem with $y\in M$ for which $xy=e.$

For example, take $a=2$ and $b=1$.

Thus, it's not a group.

Answer 2

Think about what is the difference between groups and monoids.

Groups have inverses to all elements

And try to exploit that.

Hope this helps :)

Answer 3

The Fact that will show it is Monoi not a Group is that " Inverse doesn't exist for all elements"

Let take matrix [3 0 0 6] And there doesn't exist inverse for this

While all othe properties like Associativity Closure Identity - I are satisfied. Hope it helped u