Answer to Question #218520 in Abstract Algebra for nabeel ahmed

Question #218520

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

(a) show that (G,∗) is a group.

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

(c) Give an automorphism of G. 


1
Expert's answer
2021-07-22T18:19:26-0400

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".


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