Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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??

share|improve this question
add comment

5 Answers 5

up vote 10 down vote accepted

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!

share|improve this answer
    
$\ddot\smile$ +1 –  B. S. Dec 30 '13 at 14:10
add comment

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 $\max\{a,2\}=-2$.

share|improve this answer
add comment

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

share|improve this answer
add comment

$*$ 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!

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.