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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??

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Okay, let's talk about it.

A group $G$ needs three things.

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

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

  3. 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!

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  • $\begingroup$ $\ddot\smile$ +1 $\endgroup$ – mrs Dec 30 '13 at 14:10
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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 $\max\{a,2\}=-2$.

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According to @Alexander's answer, it is a monoid instead.

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$*$ 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!

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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 $(-2)*(2^{-1})*(2)=-2$)

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