Question

Let G be the set of positive real numbers except 1. Define α∗β = α^nβ. Then:

(a) show that (G,∗) is a group.

(b) Is G abelian? if not, find its center.

(c) Give an automorphism of G. 

Answer 1

$\text{Let $\alpha,\beta,\gamma \in G$}\\ \text{Consider }\\ α∗β=α^{ln⁡β} \\ \text{Clearly, α∗β is positive since the power of any positive real number is always positive. } \\ \text{Next, suppose α∗β=1. Then we have}\\ \alpha^{\ln \beta}=1=\alpha^0 \Rightarrow \ln \beta =0 \\ \Rightarrow \beta=1\\ \text{which is a contradiction. Hence $\alpha * \beta \in G$} \\ \text{Next, consider }\\ \alpha * (\beta * \gamma)=\alpha^{\ln\left(\beta^{\ln\gamma}\right)}=\alpha^{(\ln\gamma)(\ln\beta)} \\ =\alpha^{(\ln\beta)(\ln\gamma)}=\left(\alpha^{(\ln\beta)}\right)^{\ln\gamma} \\=(\alpha * \beta) * \gamma\\ \text{Next, consider}\\ \alpha * e=\alpha^{\ln e}=\alpha^1=\alpha\\ \text{Similarly, $e*\alpha=e^{\ln\alpha}=\alpha$.} \\ \text{Finally, consider }\\ \alpha*e^{\frac{1}{\ln\alpha}}=\alpha^{\ln{(e^{\frac{1}{\ln\alpha}})}}=\alpha^{\frac{1}{\ln\alpha}} \\=\alpha^{log_{\alpha}e}=e \\ \text{Similarly, }\\ e^{\frac{1}{\ln\alpha}}*\alpha=\left(e^{\frac{1}{\ln\alpha}}\right)^{\ln\alpha}=e^{\frac{\ln\alpha}{\ln\alpha}} \\=e^1=e \\ \text{So the inverse of $\alpha \in G$ is $e^{\frac{1}{\ln\alpha}}$}​. \\ \text{Since G satisfies all the conditions of a group, we therefore conclude that G is a group.} \\ \text{Consider $e^{(\ln\alpha)(\ln\beta)}$. We can write it as} \\ e^{(\ln\alpha)(\ln\beta)}=\left(e^{(\ln\alpha)}\right)^{(\ln\beta)}=\alpha^{\ln\beta} \\=\alpha*\beta\\ \text{and} \\ e^{(\ln\alpha)(\ln\beta)}=e^{(\ln\beta)(\ln\alpha)} \\=\left(e^{(\ln\beta)}\right)^{(\ln\alpha)}=\beta^{\ln\alpha}=\beta*\alpha \\ \text{So $G$ is abelian.}\\ \text{Define a function $i:G\rightarrow G$ such that $i(g)=g\, \forall g\in G$}\\ \text{Let $g,h\in G$. Suppose $i(g)=i(h)$. Then $g=h$. So iii is one-to-one.} \\ \text{Suppose $g\in G$ . Then $g=i(g)$. So iii is onto.} \text{Finally, consider}\\ i(g)*i(h)=g*h=i(g*h)\\ \text{Hence iii is an isomorphism.} \\ \text{Thus we have an automorphism of G.}$