Groups of Automorphisms - remarks by Anderson and Feil ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 24: Abstract Groups ... ...

I need some help in understanding some claims in Chapter 24 by Anderson and Feil ... ...Anderson and Feil claim that $$\text{Aut}( \mathbb{R} )$$ is the trivial group ...

This surprises me a great deal as it seems to be saying that the number of permutations of the real number is $$1$$ ...! It seems to me that this should be an infinite number ... but obviously my intuition is way out ... ! Can someone please explain to me why $$\text{Aut}( \mathbb{R} )$$ is the trivial group ...?Peter
NOTE: Anderson and Feil also claim that the groups $$\text{Aut}( \mathbb{R} )$$ and $$\text{Aut}( \mathbb{R} )$$ are the trivial group ... but how can this be ...

 

[Euge]

Hi Peter,

You're confusing permutation with field automorphism. Permutations on a set $S$ are bijections from $S$ onto itself, whereas automorphisms of a field $F$ are bijections that preseve addition and multiplication in $F$. The dilation map $\Bbb R\to \Bbb R$, $x\mapsto 2x$ is a permuation on $\Bbb R$, but not a field automorphism since it does not map $1$ to $1$.

Let $\sigma \in \operatorname{Aut}(\Bbb R)$. Use induction arguments to show that $\sigma(q) = q$ for every rational number $q$. Then prove that $\sigma$ is a continuous. By density of the rationals in the reals, it will follow that $\sigma$ is the identity.

 

[Math Amateur]

Hi Peter,

You're confusing permutation with field automorphism. Permutations on a set $S$ are bijections from $S$ onto itself, whereas automorphisms of a field $F$ are bijections that preseve addition and multiplication in $F$. The dilation map $\Bbb R\to \Bbb R$, $x\mapsto 2x$ is a permuation on $\Bbb R$, but not a field automorphism since it does not map $1$ to $1$.

Let $\sigma \in \operatorname{Aut}(\Bbb R)$. Use induction arguments to show that $\sigma(q) = q$ for every rational number $q$. Then prove that $\sigma$ is a continuous. By density of the rationals in the reals, it will follow that $\sigma$ is the identity.

Thanks for the help Euge ... I think i can establish that an automorphism $$f $$ on $$\mathbb{R}$$ is fixed on the rational numbers $$\mathbb{Q}$$ ... ...
Suppose $$f$$ is an automorphism on $$\mathbb{R}$$ ...

Then $$f(n) = n$$ and $$f(1/m) = 1/m$$ for all $$m, n \in \mathbb{Z}$$ ...... ... to show $$f(n) = n$$ ... ...

$$f(1) = 1$$ ... ... property of automorphism$$f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$$$$f(3) = f( 1 + 2 ) = f(1) + f(2) = 1 + 2 = 3$$... and so on ... so $$f(n) = n$$
... ... to show $$f(1/m) = 1/m$$ ... ...$$f(1) = f(m \cdot m^{-1} ) = f(m) \cdot f( m^{-1} ) = 1$$$$\Longrightarrow f( m^{-1} ) = 1 / f(m) = 1/m$$ ...... that is ... $$f ( 1/m ) = 1/m$$ ... ...
The case of $$-n$$ and/or $$-m$$ is covered by the fact that $$f( -a) = - f(a) $$ for all $$a \in \mathbb{R}$$ ... ... So ... $$f$$ is fixed on the rational numbers $$\mathbb{Q}$$ ...Is that correct?Not sure how to proceed ... can you help further ...

Peter

 

[Euge]

You're almost there. Rational numbers are generally in the form $n/m$ where $n$ and $m$ are integers with $m \neq 0$ (they're not just in the form $n/1$ or $1/m$). So the missing piece of your argument is establishing $f(n/m) = n/m$ for all such $n$ and $m$.

To proceed after that point, show first that $f(x) > 0$ for all $x > 0$, then prove $\sigma$ is strictly increasing. This will help in constructing a (short) $\epsilon-\delta$ argument to prove $\sigma$ is continuous.

 

[Math Amateur]

You're almost there. Rational numbers are generally in the form $n/m$ where $n$ and $m$ are integers with $m /neq 0$ (they're not just in the form $n/1$ or $1/m$). So the missing piece of your argument is establishing $f(n/m) = n/m$ for all such $n$ and $m$.

To proceed after that point, show first that $\sigma(x) > 0$ for all $x > 0$, then prove $\sigma$ is strictly increasing. This will help in constructing a (short) $\epsilon-\delta$ argument to prove $\sigma$ is continuous.

Yes, didn't quite finish ...

We have $$f(n/m) = f(n \cdot 1/m ) = f(n) \cdot f(1/m) = n \cdot 1/m = n/m$$ ... using multiplicative property of automorphism ...

Peter

 

[Euge]

Yes, didn't quite finish ...

We have $$f(n/m) = f(n \cdot 1/m ) = f(n) \cdot f(1/m) = n \cdot 1/m = n/m$$ ... using multiplicative property of automorphism ...

Peter

Did you also prove that $f$ is continuous?