Is $\{-2,2\}$ a group under $a\star b=\max\{a,b\}$? Lets say $G={-2,2}$ and $a*b=\text{max}\{a,b\}$.
I need to check if this is a group
and if it does than is it abelian or not and finite or not.
Well... first, I'm not sure if this is a group.  for $-2,2$ I'll always get the same result so can I say there is identity number??
 A: There is an identity (with respect to the defined operation): it is $-2$, because we have $\max\{-2,a\}=a$ for any $a \in G$.
But still, $(G, *)$ is not a group. The law that is violated here is the existence of inverse elements: $2$ has no inverse, i.e. there is no such $a \in G$ that $\max\{a,2\}=-2$.
A: According to @Alexander's answer, it is a monoid instead.
A: $*$ is closed on $\{-2, 2\}$, and is well-defined.
Associative: yes
Identity: yes, we have $-2$ is an identity (with respect $*$): since $\max\{-2,g\}=g$ for any $g \in G$.
Closed under Inversion: NO: $\;2 \in G$ has no inverse, i.e. there is no such $g \in G$ that $\max\{g,2\}=-2$.
The failure of any one of the above conditions negates the prospect of $(G, *)$ being a group. Since closure under inversion fails, $(G, *)$ fails to be a group.

It's perhaps beside the point, but $*$ is commutative on $G$, and the group is finite, clearly (exactly 2 elements in $G$), so you actually have a finite, abelian monoid!
A: This is not a group because there is no $c$ such that $c*2=-2$. 
(If $G$ were a group, then we know that $(-2)*(2^{-1})$ would satisfy the property $(-2)*(2^{-1})*(2)=-2$)
A: Okay, let's talk about it.
A group $G$ needs three things.


*

*Identity:  There must be an element $e\in G$ so that $e\star g=g\star e=g$ for all $g\in G$.

*Inverses:  For each $g\in G$, there must be an element $g^{-1}\in G$ so that $g^{-1}\star g=g\star g^{-1}=e$.

*Associativity:  For $a,b,c\in G$, $a\star(b\star c)=(a\star b)\star c$.
Your example satisfies $1$ and $3$: $-2$ is an identity, and the operation is clearly associative.  However, there is no inverse for $2$.  So $G$ is not a group.  It is a monoid, though!
