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G= $Q^+$ (Rational numbers diffrent from zero) $a*b = ab/2$ I already proved this is a group now I need to prove or disprove that it is abelian and or finite group.

For abelian - from what I understad must be that for a,b -> $ab=ba$ => $a*b=b*a$ and this is easy to show. ( am I right???)

For finite - From what I saw I need to find the order but I don't really understood how to do so...

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You do know that the set $\mathbb Q\setminus\{0\}$ is infinite, don't you? –  Hagen von Eitzen Dec 26 '12 at 20:23
    
Yes but In general to find if a group is finite or not - I need to find the order? –  baaa12 Dec 26 '12 at 20:24
    
Yes, but the order is just the cardinality of the underlying set, isn't it? –  Hagen von Eitzen Dec 27 '12 at 9:15

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up vote 4 down vote accepted

It is correct to say that the group is abelian, and the method you give is correct (and it is easy, as you say. To show that you know exactly why it's true it is probably worth stating explicitly that the given operation is commutative because multiplication is as well as showing that the algebra works)

The group is infinite, and this is true because $\mathbb{Q}$ is infinite. Any more is unnecessary. However, we could approach the problem in another way. The order of a finite group $G$ is the number of elements in it. For each $g \in G$ there is some $n \in N$ such that $g^n = e$ where $e$ is the identity (if not then the group is infinite) and we call the least such $n$ the order of the element. Hence, if you can find some element for which no power will be the identity, you can prove that the group is infinite.

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thank you very much. you make it very clear for me now! –  baaa12 Dec 26 '12 at 20:34
    
@baaa12 no problem, happy to help! –  Tom Oldfield Dec 26 '12 at 20:38

(The set of all non-zero rationals is often denoted, instead, by $\mathbb{Q}^*$, but I'll stick with your notation.)

Abelian ?:

Let $a, b \in \mathbb{Q}^+$

Then $\;\;a*b = \dfrac {ab}{2} = \dfrac {ba}{2} = b * a$

$\quad\quad\quad$(...since "normal" multiplication of two rational numbers is commutative: $a\cdot b = b\cdot a$.)

Hence, the group is abelian.

The order of this group, having showed it is a group, is the order of $\mathbb{Q}^+ = |\mathbb{Q}^+| =|\mathbb{Z}|$, both of which you know are infinite.

If the order were some finite $n \in \mathbb{N}$ (which we'll simply call "$n$"), then for every element $g \in \mathbb{Q}^+$, then it would have to be true that $g^n = e$, where $e$ is the identity for $*$ in $\mathbb{Q}^+$.

Clearly, no such number $n$ exists so that $g^n = e$ is true for all elements in $\mathbb{Q}^+$. Hence, the order of your group cannot be finite, and therefore must be infinite.

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If you showed that algebric structure is a group, it is not finite. Being abelian is inherited form $\mathbb Q^*$ for your group.

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Simply summarized and to the point! –  amWhy Feb 8 '13 at 0:53

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