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Let $G=\{x \in \mathbb R \mid x>0 , x\neq 1 \}$. Define $*$ operation on $G$ by $a*b:=a^{\ln(b)}$ for all $a,b \in G$. Prove that $G$ is an abelian group under the operation $*$.

I know that definition of abelian group is like this

A group $G$ is said to be abelian if $a*b=b*a$ for all $a,b \in G$.

But how $a^{\ln b}$ should be equal to $b^{\ln(a)}$? I did not understand this and please help me, if $a*b=a^{\ln(b)}$ then by the same logic is not $b*a=b^{\ln(a)}$?if so how are they equal to each other? Thanks a lot.

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$a^{\ln b}=\exp(\ln b\ln a)$ should help. – Davide Giraudo Dec 12 '11 at 10:30
so it means that we will get equation like this exp(lnblna)=exp(lnalnb)? – dato datuashvili Dec 12 '11 at 10:37
Yes, so here, the most difficult is to show that $G$ is in fact a group. – Davide Giraudo Dec 12 '11 at 10:40
thanks a lot @Davide Giraudo – dato datuashvili Dec 12 '11 at 10:44
up vote 6 down vote accepted

First, we show that $G$ is fact a group. Since $a*b=\exp(\ln b\ln a)>0$, and $\ln b\ln a\neq 0$, $a*b\in G$. Let $a,b,c\in G$. Then $$(a*b)*c=\exp(\ln b\ln a)*c=\exp(\ln c\ln \exp(\ln b\ln a) )=\exp(\ln c\ln b\ln a)$$ and $$a*(b*c)=a*\exp(\ln c\ln b)=\exp(\ln\exp(\ln c\ln b)\ln a)=\exp(\ln c\ln b\ln a)$$ so $*$ is associative.

Check that $e$ (the number $e\approx 2.7...$) is the neutral element, and if $a\in G$ then $\exp\frac 1{\ln a}>0,\neq 1$ is the inverse of $a$.

So $(G,*)$ is a group.

Since for all $a,b\in G$, $a*b=\exp(\ln b\ln a)=\exp(\ln a\ln b)=b*a$, this group is abelian.

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i am introductory in group theory and would like to say that it was very clearly explained – dato datuashvili Dec 12 '11 at 11:03

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