Question

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

Let $\alpha,\beta,\gamma \in G$. Consider

$$\alpha * \beta = \alpha^{\ln \beta}$$

Clearly, $\alpha * \beta$ is positive since the power of any positive real number is always positive.

Next, suppose $\alpha * \beta =1$. Then we have

$$\alpha^{\ln \beta}=1=\alpha^0 \Rightarrow \ln \beta =0 \\ \Rightarrow \beta=1$$

which is a contradiction. Hence $\alpha * \beta \in G$.

Next, consider

$$\alpha * (\beta * \gamma)=\alpha^{\ln\left(\beta^{\ln\gamma}\right)}=\alpha^{(\ln\gamma)(\ln\beta)} \\ =\alpha^{(\ln\beta)(\ln\gamma)}=\left(\alpha^{(\ln\beta)}\right)^{\ln\gamma} \\=(\alpha * \beta) * \gamma$$

Next, consider

$$\alpha * e=\alpha^{\ln e}=\alpha^1=\alpha$$

Similarly, $e*\alpha=e^{\ln\alpha}=\alpha$.

Finally, consider

$$\alpha*e^{\frac{1}{\ln\alpha}}=\alpha^{\ln{(e^{\frac{1}{\ln\alpha}})}}=\alpha^{\frac{1}{\ln\alpha}} \\=\alpha^{log_{\alpha}e}=e$$

Similarly,

$$e^{\frac{1}{\ln\alpha}}*\alpha=\left(e^{\frac{1}{\ln\alpha}}\right)^{\ln\alpha}=e^{\frac{\ln\alpha}{\ln\alpha}} \\=e^1=e$$

So the inverse of $\alpha$ in $G$ is $e^{\frac{1}{\ln\alpha}}$.

Since $G$ satisfies all the conditions of a group, we therefore conclude that $G$ is a group.

Consider $e^{(\ln\alpha)(\ln\beta)}$. We can write it as

$$e^{(\ln\alpha)(\ln\beta)}=\left(e^{(\ln\alpha)}\right)^{(\ln\beta)}=\alpha^{\ln\beta} \\=\alpha*\beta$$

and

$$e^{(\ln\alpha)(\ln\beta)}=e^{(\ln\beta)(\ln\alpha)} \\=\left(e^{(\ln\beta)}\right)^{(\ln\alpha)}=\beta^{\ln\alpha}=\beta*\alpha$$

So $G$ is abelian.

Define a function $i:G\rightarrow G$ such that $i(g)=g$ for all $g\in G$.

Let $g,h\in G$ . Suppose $i(g)=i(h)$. Then $g=h$. So $i$ is one-to-one.

Suppose $g\in G$ . Then $g=i(g)$. So $i$ is onto.

Finally, consider

$$i(g)*i(h)=g*h=i(g*h)$$

Hence $i$ is an isomorphism.

Thus we have an automorphism of $G$.